Question

The sum of 3rd and 11th terms of an arithmetic progression is 34 .find the sum of 13 term of the series

Answer 1

Step-by-step explanation:

follow the above steps

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Given,

$\tt{\rightarrow T_{3}+T_{11}=34}}$

As we know that :-

$\tt{\rightarrow T_{n}=a+(n-1)d}$

a + 2d + a + 10d = 34

2a + 2d + 10d = 34

2a + 12d = 34

${\boxed{\sf\:{Taking\;common\;we\;get :-}}}$

2(a + 6d) = 34

$\tt{\rightarrow (a+6d)=\dfrac{34}{2}}$

a + 6d = 17

Hence,

a = 17 - 6d ...... (1)

Now,

$\tt{\rightarrow S_{n}=\dfrac{n}{2}[2a+(n-1)d]}$

$\tt{\rightarrow S_{13}=\dfrac{13}{2}[2(17-6d)+(13-1)d]}$

$\tt{\rightarrow S_{13}=6.5(34-12d+12d)}$

$\tt{\rightarrow S_{13}=6.5\times 34}$

= 221

Hence,

$\Large{\boxed{\sf\:{Sum\;of\;1st\;13\;terms= 221}}}$