Question

If the 19th term of an AP is 39, find the sum of its first 37 terms​

Answer 1

GIVEN :–

• 19th term of an AP = 39

TO FIND :–

• Sum of first 37 terms = ?

SOLUTION :–

• If first term of A.P. is 'a' and common difference is d , then nth term –

$\\ \longrightarrow \: \: \large { \boxed{ \sf T_{n} = a + (n - 1)d}}$

• According to the first condition –

$\\ \implies \sf T_{19} = a + (19 - 1)d$

$\\ \implies \sf 39 = a + 18d$

$\\ \implies \sf a + 18d = 39 \: \: - - - eq.(1)$

• We know that Sum of n terms –

$\\ \longrightarrow \: \: \large { \boxed{ \sf S_{n} = \dfrac{n}{2}[2a + (n - 1)d]}}$

• So that –

$\\ \implies \sf S_{37} = \dfrac{37}{2}[2a + (37 - 1)d]$

$\\ \implies \sf S_{37} = \dfrac{37}{2}[2a + 36d]$

$\\ \implies \sf S_{37} = \dfrac{37}{ \cancel2}[ \cancel2(a + 18d)]$

$\\ \implies \sf S_{37} = 37(a + 18d)$

• Using eq.(1) –

$\\ \implies \sf S_{37} = 37(39)$

$\\ \implies \large { \boxed{ \sf S_{37} = 1,443}}$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

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

• 19th term of an AP is 39.

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

• Sum of its first 37th terms

${\bf{\blue{\underline{Formula \:Used:}}}}$

$\dagger \: \boxed{\sf{ a_{n} = a + (n - 1)d}}$

$\dagger \: \boxed{\sf{ S_{n} = \frac{n}{2} 2a+ (n - 1)d}}$

${\bf{\blue{\underline{Now:}}}}$

$: \implies{\sf{ a_{19} = a + (19 - 1)d }}$

$: \implies{\sf{ 39 = a + (19 - 1)d }}$

$: \implies{\sf{ 39 = a + 18d }}$

__________________________________

$: \implies{\sf{ s_{37} = \frac{37}{2}[ 2a + (37 - 1)d] }}$

$: \implies{\sf{ s_{37} = \frac{37}{2} [2a + 36d ]}}$

$: \implies{\sf{ s_{37} = \frac{37}{2} 2(a + 18d }}$

$: \implies{\sf{ s_{37} = 37( a + 18d )}}$

$: \implies{\sf{ s_{37} = 37( 39 )}}$

$: \implies{\sf{ s_{37} = 1443}}$