Question

If the sum of first 7 terms of an AP is 119 and that of the first 17 terms is 714,find the sum of its first n terms.

Answer 1

HERE,
according to question,
CASE 1 
S7=n/2[2a+(n-1)d]
119=n/2[2a+(6)d]
238=14a+42d
119=7a+21d 

CASE-2
S17=n/2[2a+(n-1)d]
714=17/2[2a+(16)d]
1428=34a+270d
714=17a+135d

therefore,
by elimination method we get ,

Answer 2

Answer:

The sum of n terms is $S_n=3n+2n^2$

Step-by-step explanation:

Given : If the sum of first 7 terms of an AP is 119 and that of the first 17 terms is 714.

To find : The sum of its first n terms?

Solution :

$S_7 = 119 \text{ and } S_{17} =714$

The sum formula is,

$S_n=\frac{n}{2}[2a+(n-1)d]$

For 7 terms,

$S_7=\frac{7}{2}[2a+(7-1)d]$

$119=\frac{7}{2}[2a+6d]$

$a+3d=17$ ......(1)

For 17 terms,

$S_{17}=\frac{17}{2}[2a+(17-1)d]$

$119=\frac{17}{2}[2a+16d]$

$a+8d=37$ .....(2)

Now, subtracting (1) from (2), we get.

$a+8d-a-3d=37-17$

$5d=20$

$d=4$

Substituting the value d = 4 in equation (2), we get.

$a+8d=37$

$a+8(4)=37$

$a=37-32$

$a=5$

Now, substitute a=5 and d=4

$S_n=\frac{n}{2}[2a+(n-1)d]$

$S_n=\frac{n}{2}[2(5)+(n-1)(4)]$

$S_n=\frac{n}{2}[10+4n-4]$

$S_n=\frac{n}{2}[6+4n]$

$S_n=n[3+2n]$

$S_n=3n+2n^2$

Therefore, The sum of n terms is $S_n=3n+2n^2$