Question

The 17th term of an AP exceeds in 10th term by 7. FIND the common difference.

Answer 1

$\huge \sf \purple{Solution.}$

Let the first term and the common difference of

the AP be a and b respectively.

According to the question,

a₁₇ = a₁₀ + 7

a + (17 – 1) d = a + (10 – 1) d + 7

∵ aₙ = a + ( n – 1) d

a + 16 d = a + 9 d + 7

16 d – 9 d = 7

7 d = 7

$d = \frac{7}{7} = 1$

Hence, the common difference is 1 .

Answer 2

Your answer is given above.