If the sum of m terms of and A.P. is the same as the sum of its n terms, Then the sum its(m+n) terms is?

M4maths Previous Puzzles

09 Jan 2014 A.P. If the sum of m terms of and A.P. is the same as the sum of its n terms, Then the sum its(m+n) terms is?: OPtion
1) m/2
2) n/2
3) (m+n)/2
4) m+(n/2)
5) mn
6) mn/2
7) 0
8) m
9) n
10)None of these; Correct Option: 7 Best Solution (33)

Hardik Patel 11 years AGO

according to question

m/2 * (2a + (m-1)d) = n/2 * (2a + (n-1)d)

cutting 2

we get

m(2a + (m-1)d) - n(2a+ (n-1)d) = 0

2am + m^2d - md -2an -n^2d +nd =0

2a(m-n) + (m^2 - n^2)d -(m - n)d =0

2a(m-n) + ( (m + n) (m- n) ) d - (m - n )d = 0

taking (m-n) common

2a + ( m + n -1) d = 0------------ 1

S m+n = m+n/2( 2a + (m+n -1)d

we know that 2a + (m+n)d is 0 from eqn. 1

therefore S m+n = 0

Like? Yes (107) | No (25)

mahima srivastava 11 years AGO

Sn=(n/2)(2a+(n-1)d) ....eq(i)
Sm=(m/2)(2a+(m-1)d) ....eq(ii)
Sm+n=((m+n)/2)(2a+(m+n-1)d) .....eq(iii)
given, Sn=Sm
therefore,
2am+dm^2-md = 2an+dn^2-nd
2a(m-n)+d(m^2-n^2-(m-n))=0
m-n(2a+d(m+n-1)=0
2a+d(m+n-1)=0 .... eq(iv)
from eq(iii) and eq(iv) we get
Sm+n=0
so, option 7

Like? Yes (54) | No (17)

Chandni Maurya 11 years AGO

ans=0
given,
sum of m terms= sum of n terms
Sm=Sn
(m/2)[2a+(m-1)d]=(n/2)[2a+(n-1)d]
m[2a+(m-1)d] = n[2a+(n-1)d]
2a(m-n)= d[n(n-1)-m(m-1)]
2a(m-n)= d[(n^2-m^2)-(n-m)
2a(m-n)= d[(n-m)(n+m)-(n-m)]
2a(m-n)= -d[(m-n)(m+n-1)]
2a= -d(m+n-1)
2a+d(m+n-1)=0---(1)
now,
sum of (m+n) terms
S(m+n)= {(m+n)/2}[2a+(m+n-1)d]
from(1)
S(m+n) ={(m+n)/2}[0] = 0

Like? Yes (28) | No (16)

ROHIT UPADHYAY 11 years AGO

m terms=m/2[2a+(m-1)d]
n terms=n/2[2a+(n-1)d]
because m terms = n terms
m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m^2d-md=2an+n^2d-nd
2a(m-n)+(m-n)(m+n)d - md + nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0
(2a-d)(m-n)=2a(m-n)- md + nd
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a ---------eq(1)
m+n terms=n/2*[2a+(m+n-1)d]
=n/2*[2a-2a]
=0

Like? Yes (22) | No (12)

Bharat Bhushan Jha 11 years AGO

let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a (equation 1)
sum of m+n terms = m=n/2*[2a+(m+n-1)d] (equation 2)
put the vaue of equation 1 to equation 2 you will get the zero

Like? Yes (12) | No (5)

Abhimanyu singh 11 years AGO

Sm=Sn

->m[2a+(m-1)d]/2=n[2a+(n-1)d]/2
->2a(m-n)+{m(m-1)-n(n-1)}d=0
->2a(m-n)+{(m^2-n^2)-(m-n)}d=0
->(m-n)[2a+(m+n-1)d]=0
->2a+(m+n-1)d=0

.'. Sm+n=(m+n)[2a+(m+n-1)d]/2
=(m+n)[0]/2
=0

Like? Yes (11) | No (5)

Puja Jha 11 years AGO

let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a (equation 1)
sum of m+n terms = m=n/2*[2a+(m+n-1)d] (equation 2)
put the value of equation 1 to equation 2 you will get the zero.

Like? Yes (14) | No (10)

Pargat Pal Singh 11 years AGO

m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
solving it, we get
(m+n-1)d= -2a
sum of m+n terms = [(m+n)/2]*[2a+(m+n-1)d] =0

Like? Yes (12) | No (8)

LOGAN WOLVERINE 11 years AGO

Like? Yes (5) | No (3)

Krishna Chaitanya 11 years AGO

Given sum of m terms of and A.P. is the same as the sum of its n terms
We know sum of n terms of AP= n/2(2a+(n-1)d)
Therefore sum of m terms =m/2(2a+(m-1)d)----(1)
sum of n terms=n/2(2a+(n-1)d)-----(2)
According to the problem (1)=(2)
m/2(2a+(m-1)d)=n/2(2a+(n-1)d)
2ma+(m2-m)d=2na+(n2-n)d
(2m-2n)a=(n2-n)d-(m2-m)d
2(m-n)a=n2d-nd-m2d+md
2a(m-n)=-d(m-n)(m+n-1)
a=-d/2(m+n-1)----(3)
From (3) we got a,sum of m+n terms=m+n/2(2(-d/2(m+n-1)+(m+n-1)d)
which is 0
Hence sum of its(m+n) terms is Zero

Like? Yes (10) | No (8)

Subham Mahanta 11 years AGO

The sum of m terms of the A.P. is = m/2(m+1)
the sum of n terms of the A.P. is = n/2(n+1)
from the question, we get,
m/2(m+1) = n/2(n+1)
=> m2+m = n2+n
=>(m2-n2)+(m-n)=0
=>(m+n)(m-n)+(m-n)=0
=>(m-n)(m+n+1)=0
=>(m+n+1)=0
=>{(m+n)/2}*(m+n+1)=0*{(m+n)/2}
=>{(m+n)/2}*(m+n+1)=0
=> the sum of (m+n) terms of the A.P.=0

Like? Yes (7) | No (5)

shivam tiwari 11 years AGO

Let AP be first term = A and common difference = D, then Question gives
SM = SN gives M(2A + MD - D) = N(2A + ND - D) whence [2A + (M+ N-1)D] = 0
hence SM+N =(M+N)[2A + (M+N-1)D]/2 =0

Like? Yes (5) | No (3)

AMIT JAISWAL 11 years AGO

here , sn=sm

m/2[2a+(m-1)d]=n/2[2a+(n-1)d]
am+(m^2-m)d/2=an+(n^2-n)d/2
2am+(m^2-m)d=2an+(n^2-n)/2
2a(m-n)=(n^2-n)d-(m^2-m)d
now adding mn both sides we get-
2a(m-n)+mn=(mn+n^2-n-m^2+m)d
2a(m-n)= (n^2+mn-n-m^2-mn+m)
2a(m-n)= n(m+n-1)d - m(m+n-1)d
2a(m-n)= (n-m)(m+n-1)d
2a = (1-m-n)d

so. s(m+n) = (m+n)/2[2a+(m+n-1)d]
= (m+n)/2[(1-m-n)d+ (m+n-1)d]
= 0

Like? Yes (7) | No (5)

RagunathKC 11 years AGO

let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d];
sum of n terms=n/2[2a+(n-1)d];
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a ....(eq1)
sum of m+n terms = (m+n)/2*[2a+(m+n-1)d].... (eq2)
put the vaue of eq1 to eq2 you will get the zero

Like? Yes (6) | No (5)

Ankit Goyal 11 years AGO

This is property of an AP

Like? Yes (13) | No (12)

Prateek Kumar 11 years AGO

S(n)=n/2[2a+(n-1)d]
S(m)=m/2[2a+(m-1)d]
S(n)=S(m)
so;on equating ,we get
[2a+(m+n-1)d]=0 ;m not equals n
S(m+n)=(m+n)/2[2a+(m+n-1)d]=0

Like? Yes (5) | No (4)

OM PRAKASH JHA 11 years AGO

Let AP be first term = a and common difference = d, then Question gives
Sm = Sn gives m(2a + md -- d) = n(2a + nd -- d) whence [2a + (m+ n -- 1)d] = 0
hence Sm+n =(m+n)[2a + (m+n--1)d]/2 =0

Like? Yes (6) | No (5)

julee verma 11 years AGO

sn=sm

m/2[2a+(m-1)d]=n/2[2a+(n-1)d]
am+(m^2-m)d/2=an+(n^2-n)d/2
2am+(m^2-m)d=2an+(n^2-n)/2
2a(m-n)=(n^2-n)d-(m^2-m)d
now adding mn both sides we get-
2a(m-n)+mn=(mn+n^2-n-m^2+m)d
2a(m-n)= (n^2+mn-n-m^2-mn+m)
2a(m-n)= n(m+n-1)d - m(m+n-1)d
2a(m-n)= (n-m)(m+n-1)d
2a = (1-m-n)d

so. s(m+n) = (m+n)/2[2a+(m+n-1)d]
= (m+n)/2[(1-m-n)d+ (m+n-1)d]
= 0

Like? Yes (2) | No (1)

Rahul gulati 11 years AGO

Like? Yes (3) | No (2)

KRISHNAN 11 years AGO

The sum of p terms of an arithmetic series is
p2(2a+(p−1)d) where a is the first term and d is the difference between each term.
We can express what m and n equal to by putting p equal to n and m respectively.

m/2 * (2a + (m-1)d) = n/2 * (2a + (n-1)d)

cutting 2

we get

m(2a + (m-1)d) - n(2a+ (n-1)d) = 0

2am + m^2d - md -2an -n^2d +nd =0

2a(m-n) + (m^2 - n^2)d -(m - n)d =0

2a(m-n) + ( (m + n) (m- n) ) d - (m - n )d = 0

taking (m-n) common

2a + ( m + n -1) d = 0------------ 1

S m+n = m+n/2( 2a + (m+n -1)d

we know that 2a + (m+n)d is 0 from eqn. 1

therefore S m+n = 0

Like? Yes (3) | No (3)

Naveen Pandey 11 years AGO

Like? Yes (4) | No (4)

Surya Narayana K 11 years AGO

Given that , Sum of m terms is equal to sum of n terms in A.P. So, m(m+1)/2 = n(n+1)/2.

=>m^2+m=n^2+n
=>m^2-n^2=n-m
=>(m+n)(m-n)=-(m-n)
=>m+n=-1

Sum of its (m+n) terms is (m+n)(m+n+1)/2.

so,Sum of its (m+n) terms is Zero.(Since m+n=-1==>m+n+1=0).

Therefore, Sum of its (m+n) terms is zero(Option 7).

Like? Yes (2) | No (2)

OM PRAKASH JHA 11 years AGO

Let AP be first term = a and common difference = d, then Question gives
Sm = Sn gives m(2a + md -- d) = n(2a + nd -- d) whence [2a + (m+ n -- 1)d] = 0
hence Sm+n =(m+n)[2a + (m+n--1)d]/2 =0

Like? Yes (4) | No (4)

vignesh 11 years AGO

let a and c be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+m square*d-md=2an+n square*d-nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
(2a-d)(m-n)+(m+n)(m-n)d=0 { (2a-d)(m-n)=2a(m-n)-md+nd }
m+n= -(2a-d)/d
(m+n)d-d= -2a
(m+n-1)d= -2a (equation 1)
sum of m+n terms = m=n/2*[2a+(m+n-1)d] (equation 2)
put the vaue of equation 1 to equation 2 you will get the zero

Like? Yes (2) | No (2)

vishal singh thapa 11 years AGO

SM={2A + (M-1)D}M/2
SN= {2B+(N-1)D}N/2
SINCE
SM=SN
{2A+D(M-1)}M/2={2B+(N-1)D}N/2
2MA+M(M-1)D=2NA+N(N-1)D
2A(M-N)=(N^2-N-M^2+M)D
-2A=(M+N-1)D ....EQ(1)
HENCE
S(M+N)={2A+(M+N-1)D}(M+N)/2
S(M+N)=0 {FROM EQ(1)}

Like? Yes (2) | No (2)

Sana 11 years AGO

let a and d be the first term and difference respectively
now, sum of m terms= m/2*[2a+(m-1)d]=n/2[2a+(n-1)d]
so, m[2a+(m-1)d]=n[2a+(n-1)d]
2am+(m*m*d)-md=2an+(n*n*d)-nd
2am+(m*m*d)-md-2an-(n*n*d)+nd
2a(m-n)+(m-n)(m+n)d-md +nd=0
2a(m-n)+(m-n)(m+n)d-d(m-n)=0
2a(m-n)+(m-n)(m+n)d-d(m-n)=0
2a+(m+n)d-d=0;
(m+n-1)d=-2a;[equation 1]
sum of m+n terms = m=(m+n)/2*[2a+(m+n-1)d] (equation 2)
put the vaue of equation 1 to equation 2 you will get the zero

Like? Yes (2) | No (2)

VIKASH RAY 11 years AGO

Like? Yes (4) | No (5)

RAKESH 11 years AGO

Sm=Sn
m/2[2a+(m-1)d]=n/2[2a+(n-1)d]
or 2a(m-n)+d(m^2-m-n^2+n)=0
or 2a(m-n)+d(m-n)(m+n-1)=0
or (m-n)[2a+(m+n-1)d]=0
or [2a+(m+n-1)d]=0 ......(1)
so S(m+n)=(m+n)/2 *[2a+(m+n-1)]=0 from eqn(1)

Like? Yes (2) | No (3)

chirag jain 11 years AGO

Sp=Sq
p/2[2a +(p-1)d]=q/2 [2a+(q-1)d]
p[2a +pd - d]=q[2a+qd - d ]
2ap + p2d - pd =2aq + q2d -qd
2ap-2aq +p2d -q2d -pd +qd =0
2a (p-q) +(p+q)(p-q)d -d(p-q)=0
(p-q)[2a + (p+q)d - d ]=0
2a + (p+q)d - d=0
2a + [(p+q)-1]d=0
hence prove that the sum of its first (p+q) terms is 0.

Like? Yes (3) | No (4)

Dinesh thatti 11 years AGO

Like? Yes (2) | No (3)

navneet shankhwar 11 years AGO

sum of m term is equal to sum of n term.
m/2[2a+(m-1)d]=n/2[2a+(n-1)d]
on solving above eq., then we get
2a+(m+n-1)d=0
sum of m+n terms is
m+n/2[2a+(m+n-1)d]=m+n/2[0]=0

Like? Yes (5) | No (6)

Rituraj Singh 11 years AGO

sum of m terms = sum of n terms
=> m/2(2a+(m-1)d) = n/2(2a+(n-1)d)

on solving this equation we get;

2a + (m+n-1)d = 0..............................(1)

therefore,
sum of m+n terms is

(m+n)/2(2a + (m+n-1)d)
using (1) we get it equal to 0(zero)

Like? Yes (2) | No (3)

NILOTPAL MRINAL 11 years AGO

n/2[2a+(n-1)d]=m/2[2a+(m-1)d]
after solving it
we get the value of m+n..
which will come 0

Like? Yes (3) | No (5)

Dimple

India

Punjab

Time: 00:01:33

Points

Maths Quotes

" MATHEMATICS IS GIVES THE CREATIVITY THAT HOW TO SOLVE ANY PROBLEM IN ANY FIELDS "

Ajay Singh

Arithmetic is where numbers fly like pigeons in and out of your head.

Carl Sandburg

Placed User Comments

M4Math helped me a lot.

Vipul Chavan 4 years ago

Thanks m4 maths for helping to get placed in several companies.
I must recommend this website for placement preparations.

yash mittal 5 years ago

If the sum of m terms of and A.P. is the same as the sum of its n terms, Then the sum its(m+n) terms

M4maths Previous Puzzles

09 Jan 2014 A.P.