A person borrows Rs.4500 and promises to pay back (without any interest) in 30 installment, each of

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16 Aug 2013 A person

A person borrows Rs.4500 and promises to pay back (without any interest) in 30 installment, each of value Rs.10 more than the last (preceding one). Find the last installment.

OPtion

1) 215
2) 225
3) 230
4) 245
5) 265
6) 295
7) 300
8) 310
9) 340
10) None of these

Solution

Let a be the first installment
Successive installment are
a, a+10, a+20
here d=10, n=number of installment =30
Solve by using A.P. Series
S=n/2(2a+(n-1)d)=4500
solve
first intallment 5, last intallment 295
option 6

Correct Option: 6 Best Solution (16)

Devendra Marghade   11 years AGO

From arithmetic sum series formula,
if a=first installment,n=30, d=10 then
n/2[2a+(n-1)10]=4500 or 15[2a+290]=300 ,a=5
and 30th term=a+29d=5+29*10=295

Like? Yes (27) | No (4)  

Tapesh mishra   11 years AGO

S=4500
n=30
d=10
s=n/2[2a+(n-1)d]
a=5
LAST TERM=5+290=295

Like? Yes (7) | No (1)  

mandeep   11 years AGO

let x be the amount of first installment
as every installment is 10 more than the last one so the 2nd installment will be (x+10)
third will be (x+20)
it forms an A.P.
their sum is
(x)+(x+10)+(x+20)+(x+30)..........
this sum is equal to 4500.
as their are 30 installments to be paid
applying the sum formula to this A.P.
S=n/2(2x+(n-1)10)
S is 4500 and n is 30
on substituting values, 'x' will be obtained as 5.
i.e. first installment is of Rs. 5
for the 30th installment(last installment),the amount he pays is determined by using the formula
x+(n-1)10 (x=50)
on putting values it gives the answer 295
so the last instmnt is Rs. 295

Like? Yes (10) | No (4)  

SHIVANI PRAKASH   11 years AGO

let 1st inst. be x
next inst.=x+10
similarly,nxt inst are:x+20,x+30......x+290
the person pays back without any interest,thus sum of all the inst.=4500
thus sum of all the inst.(in A.P.):
(30/2)*(2x+290)=4500
x=5,thus last installment=x+290=295

Like? Yes (3) | No   

Balakrishnan M   11 years AGO

Sn= (n/2)[2a+(n-1)d]
4500=(30/2)[2a+(30-1)10]
Solve this we get,
a=5
Last term=a+(n-1)d
=5+(30-1)10
=5+(29)10
=5+290
=295

Like? Yes (3) | No   

Bingumalla Abinay   11 years AGO

4500=30/2(2a+290)
a=5
therefore Snth term =a+(n-1)d=5+290=295

Like? Yes (3) | No (1)  

siddhartha   11 years AGO

let the first installment is X

then next 2nd installment is X+10
and then next 3rd installment X+20

and so on... last 30th installment is X+290

so according to prob..

4500 = X+(X+10)+(X+20)+(X+30)......+(X+290)
= 30X + 10*(1+2+3+..+29)
= 30X + 4350

so X=5 rs which is first installment

last installment = X+290 = 295 RS

Like? Yes (2) | No   

richa ahuja   11 years AGO

sn=4500
n=30
d=10
sn=n/2[2a+(n-1*d)]
putting values of sn,d,n we get
a=5
l=a+(n-1)*d
l=5+(29)*10=295

Like? Yes (2) | No   

Aarush Singh   11 years AGO

starting from Rs.5 and adding Rs.10 we get the last installment Rs.295 which will payback rs4500 (without any interest)

Like? Yes (2) | No   

Suman Saurabh   11 years AGO

Let first installment=a
Total no. of installment=30
Last installment=a+(30-1)*10=a+290
4500=15(2*a+290)
a=5
so Last installment=295

Like? Yes (3) | No (1)  

vishnu vardhan   11 years AGO

given,4500 to be payed in 30 instalments with out any interest andfor every next instalment there is increase in price of 10.
therefore,let first installment price be 'x'
then total instalments will be in the order,
=>x+(x+10)+(x+10+10)+(x+10+10+10)+............
=>x+(x+10)+(x+20)+.........+(x+290)=4500.
=>30x+(10+20+30+.......+290)=4500.
=>3x+(1+2+3+......+29)=450.
=>3x+435=450.
=>3x=15.
=>x=5.
therefore,last installment=290+x,=>290+5=295 rupees.

Like? Yes (2) | No (1)  

sudipta sadhukhan   11 years AGO

lets assume first installment Rs. x.
it follows a AP series with c.d. 10.
so,
30/2[2*x+(30-1)*10]=4500
solving we get x=5
so ans is 5+(30-1)*10=295.

Like? Yes (1) | No   

ShAg   11 years AGO

this is in Arithmetic Progression.
let first installment is of Rs.x, 30 installments are to be paid with increment of Rs.10 in the previous installment.
So, in our formula L = a + (n-1)d, to find the last installment, a=x, n=30, and d=10
hence our eqn comes out to be:
L = x + (30-1)10
=> L = x + 290
for finding the sum of an Arithmetic Progression, formula is: S = n(a+L)/2
so our eqn comes out to be:
4500 = 30(x+x+290)/2
=> x=5
Putting x in L = x + 290, we get the answer:
L = 5 + 290 = Rs.295

Like? Yes (1) | No   

Chandni Maurya   11 years AGO

ans=(6). 295
let x be the first installment,
x+x+10+x+20+....+x+290=4500
=> 30x + 29/2(2*10+(29-1)10))=4500
=> 30x + 4350 =4500
=> 30x = 150
=> x = 5 (1st installment)
=> 5 + 290 = 295 (last installment)

Like? Yes (1) | No (1)  

paramesh   11 years AGO

let us assume that

in first installmenthe pays - x
in second - x+10
in third - x+20
similarly..
in 30th instlmnt -x+290

By adding all..
we will get 30x+4350=4500
from above equation x=5 ruppes
we know that in 30th installment he should pay X+290 ruppes..here x=5..So answer is 290+5=295..

Like? Yes (1) | No (1)  

GnanaMeena   11 years AGO

295+285+275+265+255+245+235+225+215+205+195+185+175+165+145+135+125+115+105+95+85+75+65+45+35+25+15+5=4500

Like? Yes (2) | No (11)