Question

find the sum of first 35 terms of the series whose pth term is p/7+2

Answer 1

Solution:-
a₁ = (1/7) + 2
= 15/7
a₂ = (2/7) + 2
= 16/7
a₃ = (3/7) + 2 
= 17/7
a₃₅ = (35/7) + 2
a₃₅ = 49/7
= 35/2 {15/7 + 49/7}
= 35/2 (64/7)
= 160
Answer.

Answer 2

Answer:

The sum of first 35 terms of the series is 160.

Step-by-step explanation:

Given : The p th term is $a_p=\frac{p}{7}+2$

To find : The sum of  first 35 terms of the series ?

Solution :

First we find the series by putting the values of p,

For first term, put p=1

$a_1=\frac{1}{7}+2=\frac{15}{7}$

Put p=2

$a_2=\frac{2}{7}+2=\frac{16}{7}$

So, The series is $\frac{15}{7},\frac{16}{7},...$

The difference of the series

$d=\frac{16}{7}-\frac{15}{7}=\frac{1}{7}$

We have to find the sum of first 35 terms i.e. n=35

Applying sum formula of an A.P is

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

Substitute the value,

$S_{35}=\frac{35}{2}(2(\frac{15}{7})+(35-1)\frac{1}{7})$

$S_{35}=\frac{35}{2}(\frac{30}{7}+\frac{34}{7})$

$S_{35}=\frac{35}{2}(\frac{64}{7})$

$S_{35}=5\times 32$

$S_{35}=160$

Therefore, The sum of first 35 terms of the series is 160.