How many terms of the series 1,4,7,11........ are needed to give the sum 715.

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17 Apr 2013 AP

How many terms of the series 1,4,7,11........ are needed to give the sum 715.

OPtion

1) 19
2) 20
3) 21
4) 22
5) 23
6) 24
7) 25
8) 26
9) 27
10) None of these

Solution

Sn= 715
then use AP Formula (Sn = n/2(2a+(n-1)d)
ans 22
option 4

Correct Option: 4 Best Solution (11)

Sahil   11 years AGO

Nice puzzle.

First correct the series to 1,4,7,10,.....
AP series having
a=1
d=3
n= ?
so that
sum =715
now
Sum, s= (n/2)*(2n-(n-1)*d)
715 = (n/2)*(2-(n-1)*3)
solving , we get
n=22
But
If series is 1,4,7,11,... then it is a different case.
After every two terms difference is increasing.
Then series will be
1,4,7,11,15,20,25,31,37,...

Like? Yes (7) | No (2)  

Anita   11 years AGO

Solved considering series as 1,4,7,10...
an AP series with
a=1
d=3

sum =715
using
s= (n/2)*(2n-(n-1)*d)
715 = (n/2)*(2-(n-1)*3)
solving , we get
n=22

Like? Yes (7) | No (4)  

Ashu   11 years AGO

715 = (x/2)*(2+(x-1)*3)
x=22

Like? Yes (5) | No (3)  

Balakrishnan M   11 years AGO

From the given AP is 1,4,7,10....
First term a=1,
common difference d=3
sum=715
sum=(n/2)[2a+(n-1)d]
715=(n/2)[2+(n-1)3]
3n^2-n-1430=0
solve this equation we get n=22
so 22 terms needed to give the sum 715.

Like? Yes (4) | No (4)  

Shoaib   11 years AGO

For the given AP,
a = 1,
d = 3 and
Sn = 715.
We have:


Sn= n/2[2a+(n-1)d]

715=n/2[2(1)+(n-1)3]

1430=n[3n-1]

1430=3n^2-n

3n^2-n-1430=0

n=22 or n=-65/3

Since n cannot be negative, so, n = 22

Like? Yes (4) | No (4)  

Ankit Pandey   11 years AGO

S=n/2[2a+(n-1)d]
=> 715 =n/2 [2*1+(22-1)*3]
On solving for n, we will get n=22.

Like? Yes (4) | No (5)  

jomon v j   11 years AGO

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

715 = n/2(2+(n-1)3)

1430 = n(3n-1)

3n²-n-1430 =0

n= 22

Like? Yes (3) | No (4)  

Rats   11 years AGO

S=(n/2)(2a+(n-1)d)
Here, a=1, d=3 and S=715
715=(n/2)(2+(n-1)3)
715=(n/2)(3n-1)
3n^2-n-1430=0
n=22

Like? Yes (3) | No (5)  

krishnapriya   11 years AGO

Given sum of the series=715
in the given series a=1,d=3
therefore, n/2[2a+(n-1)d]=715
(ie)n/2[2+(n-1)3]=715
Solving the above eqn.
3n^2-n-1430=0
n=22

Like? Yes (2) | No (4)  

Vinay Sipani    11 years AGO

Sn = (n/2) ( 2a + (n - 1)d )

=> 715 * 2 = n ( 3n - 1)
=> n = 22

Like? Yes (2) | No (4)  

ANCHAL MANGAL   11 years AGO

by using the formula
s=n/2[2a+(n-1)d]
a=1,d=3

Like? Yes (3) | No (6)