# If the 8th term of an AP is 31 and its 15th term is 16 more than the 11th term, find the AP.

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If the 8th term of an AP is 31 and its 15th term is 16 more than the 11th term, find the AP.

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

We know that,

nth term of AP is an = a + (n – 1) d.

Given that,

8th term of AP is a8 = 31

⇒ a + (8 – 1) d = 31

⇒ a + 7d = 31 … (1)

Also,

Given that,

15th term of AP is 16 more than the 11th term of AP.

i.e., a15 = 16 + a11

⇒ a + 14d = 16 + a + 10d

⇒ 14d − 10d = 16

⇒ d = $\frac{16}{4}$ = 4.

From equation (1),

a + 7 × 4 = 31

(∵d = 4)

⇒ a = 31 – 28 = 3.

Hence,

The first term of AP is a1 = a = 3.

The second term of AP is a2 = a + d

= 3 + 4 = 7.

The third term of AP is a3 = a + 2d

= 3 + 8

= 11.

Hence,

The required AP is 3, 7, 11, ……… .

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