Question

find the common difference of an ap whose first term is 4 last term is 49 and sum of all of its term is 265

Answer 1

Answer:

here is the answer

let first term is a and difference is and total terms are n then

a=4

a+(n-1) d=49

(n-1) d=49-4=45--->1

sum of n terms is 265

then

n×{2a+(n-1) d}=265×2---->2

put value in eq. 2

then we get

n×(8+45) =530

n×53=530

n=10

then

(n-1) d=45

9×d=45

d=5

common difference is 5

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Answer 2

ANSWER :

BY USING THE FORMULA:

Sn = n/2(a + l)

265 = n/2(4 + 49)

530 = 53n

THUS,

n = 10

PUTTING THE VALUE IN FORMULA :

Sn = n/2(2a + [n - 1]d)

265 = 10/2(8 + 9d)

265 = 5(8 + 9d)

265 = 40 + 45d

225 = 45d

5 = d