Question

find the sum of the first 25 terms of the AP 21, 32, 43...​

Answer 1

a= 21 , d=32-21=> 11, n=25

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

s= $\frac{25}{2}$[2(21)+(25-1)11]

s= $\frac{25}{2}$[42+264]

s= $\frac{25}{2}$[306]

s= 25[153]

s= 3825

Answer 2

$\huge\mathbb\orange{ANSWER!!!!}$

Given -

• a = 21

d = a² - a¹

• d = 32 - 21

• d = 11

• n = 25

we know the formula that,

• sn = n/2{ 2a + (n-1)d}

Now,

• sn = 25/2 {2 × 21 + (25 - 1) 11}

• sn = 25/2 { 42 + 24 × 11}

• sn = 25/2 (42 + 264)

• sn = 25/2 × 306

• sn = 25 × 153

• sn = 3825

Hence,

• Sum of 25th term of the AP is 3825.