Question

the first and last of an ap are 7 and 49 respectively if sum of all its term is 420 find common difference

Answer 1

Given first term = a = 7
Last term = l = 49
Sn = 420
⇒ n/2[ a+l ] = 420
⇒ n/2[ 7+49 ] = 420
⇒ n/2[ 56 ] = 420
⇒ n/2 = 420/56
⇒ n/2 = 7.5
⇒ n = 7.5 ×2
∴ n = 15
We know that Sum of these 15 terms = 420
⇒ n/2[ 2a+(n-1)d ] = 420
⇒ 15/2[ 2(7) + (15-1)d ] = 420
⇒ 15/2[ 14 + 14d ] = 420
⇒ 15[ 14 + 14d ] = 420 ×2
⇒ 15[ 14 + 14d ] = 840
⇒ 14 + 14d = 840/15
⇒ 14 + 14d = 56
⇒ 14d = 56 - 14
⇒ 14d = 42
⇒ d = 42/14
∴ d = 3
∴ The common difference is 3

Answer 2

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