Question

find the sum of first 76 terms of an AP whose second and third term are 12 and 20 respectively​

Answer 1

Answer; 23712

Step-by-step explanation:

a+2d-a+d=d

20-12=8

d=8 and a=4

na+ n(n+1)/2d=23712 on substitution

n=76

Answer 2

$\huge\tt{\underline{ \underline{Answer :}}}$

$\underline{ \tt{Given } : }$

$\tt{a2 = 12} \\ \tt{a3 = 20}$

$\tt{S76 = \: \: ?}$

$\tt{We \: \: can \: \: write : }$

$\tt{a2 = a + 1d \: = 12} - - - - - (1)$

$\tt{a3 = a + 2d \: = 20} - - - - - (2)$

$\tt{Solving \: \: equations \: \: (1) \: \: and \: \: (2)} :$

$\tt{a + 1d = 12 } \: \: \: - \\ \tt \underline{a + 2d = 20} \: \: \: \: \: \: \: \\ \tt{ 0 - 1d = - 8} \: \: \: \: \: \:$

$\tt{ d = 8}$

$\tt{Substituting \: \: \: d=8 \: \: in \: \: (1)}$

$\tt{a + (8) = 12} \\ \tt{ a = 12 - 8} \: \: \: \\ \tt{ a = 4 }\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\boxed{ \tt{a = 4 \: \: \: \: \: \: d = 8}}$

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

$\tt{S76= \frac{76}{2} [2(4) + (75)8]}$

$\tt{ = 38[8 +60 0]}$

$\tt{ = 38[608]}$

$= \underline{ \underline{ \tt{ 23104}}}$

$\tt{Hope \: \: this \: \: helps \: \: you \: !}$