Sum

The 10^{th} term of an A.P. is 52 and 16^{th} term is 82. Find the 32nd term and the general term

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#### Solution

Let a be the first term and d be the common difference of the given A.P. Let the A.P. be

a_{1} , a_{2} , a_{3} , ….. an, ……

It is given that a_{10} = 52 and a_{16} = 82

⇒ a + (10 – 1) d = 52 and a + (16 – 1) d = 82

⇒ a + 9d = 52 ….(i) and, a + 15d = 82 ….(ii)

Subtracting equation (ii) from equation (i), we get

–6d = – 30 ⇒ d = 5

Putting d = 5 in equation (i), we get

a + 45 = 52 ⇒ a = 7

∴ a_{32} = a + (32 – 1) d = 7 + 31 × 5 = 162

and, an = a + (n – 1) d = 7 (n – 1) × 5 = 5n + 2.

Hence a_{32} = 162 and an = 5n + 2

Concept: Arithmetic Progression

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