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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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