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

1.	The 17th term of an AP exceeds in 10th term by 7. FIND the common difference.
Answer» $\huge \sf \purple{Solution.}$ 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 .

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

Answer»

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

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 .

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

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