Question

hi guys ...............................please helpme......................................................................... Tomorrow am going to exam...................fast

$the \: 17th \: term \: of \: an \: ap \: exceeds \: its \: 10th \: term \: by \: 7.find \: the \: common \: difference$

Answer 1

According to the question,

17th term of an A. P., A17= a + 16 d

10th term of an A. P., A 10 = a + 9d

A17 - A10 = 7

a + 16d - a - 9d = 7

5d = 7

d = 7 / 5

Common difference, d = 7/5

Answer 2

Answer:

1

Step-by-step explanation:

Let a be the first term and d be the common difference.

nth term of an AP an = a + (n - 1) * d.

(i) 17th term:

a₁₇ = a + (17 - 1) * d

    = a + 16d.

(ii) 10th term:

a₁₀ = a + (10 - 1) * d

    = a + 9d.

Now,

17th term exceeds its 10th term by 7.

a₁₇ = a₁₀ + 7

⇒ a + 16d = a + 9d + 7

⇒ a + 16d - a - 9d = 7

⇒ 7d = 7

⇒ d = 1.

Therefore, Common difference is 1.

Hope it helps!