Question

If the 2nd and
and 13th term st AP р
are - 5 and 39
Find the
sum
ot 아 the first 35
terms
of the AP.
.​

Answer 1

Answer:

$\bigstar{\bold{S_{35}=2065}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• 2nd term of the A.P (a₂) = -5
• 13th term of the A.P (a₁₃) = 39

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Sum of first 35 terms of the A.P

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ First we have to find the common difference of the A.P

→ Common difference of the A.P is given by

  $d=\dfrac{a_m-a_n}{m-n}$

where $a_m$ = 39, $a_n$ = -5 , m = 13, n = 2

→ Substituting the datas we get,

  $d=\dfrac{39+5}{13-2}$

 $d=\dfrac{44}{11}$

d = 4

→ Hence common difference of the A.P is 4

→ Now we need to find sum of the first 35 terms

→ Sum of n terms of an A.P is given by

  $S_n=\dfrac{n}{2}(2a_1+(n-1)\times d)$

 where n = 35, a₁ = a₂ - d = -5 - 4 = -9

→ Substituting the datas,

  $S_{35}=\dfrac{35}{2}(2\times-9 + (35-1)\times 4)$

  $S_{35} = 17.5 (-18 + 136)$

 $S_{35} = 17.5 \times 118$

 S₃₅ = 2065

→ Hence sum of first 35 terms of an A.P is 2065

  $\boxed{\bold{S_{35}=2065}}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ The sum of n terms of an A.P is given by

  $S_n=\dfrac{n}{2}(a_1+a_n)$

  $S_n=\dfrac{n}{2}(2a_1+(n-1)\times d)$