Question

The 15th term of an A.P exceeds it's 10th term by 25, the common difference is ?

Answer 1

Answer:

a15=a10 + 25

a+14d=a+9d+25

a+14d-a-9d=25

5d=25

d=25/5

d=5

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Answer 2

Solution :-

Let first term of given AP is a and common difference is d .

So,

→ T(n) = a + (n - 1)d

then,

→ T(10) = a + 9d --------- (1)

→ T(15) = a + 14d ----------- (2)

A/q,

→ (2) - (1) = 25

→ (a + 14d) - (a + 9d) = 25

→ a - a + 14d - 9d = 25

→ 5d = 25

→ d = 5 (Ans.)

Hence, the common difference is equal to 5 .

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