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If the sum of 1st n terms of an AP is \(\frac{1}2\) (3n2 + 7n) then find its nth term and hence, write its 20th term. |
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Answer» Given that the sum of first nth term of AP is Sn = \(\frac{1}2\) (3n2 + 7n). ∴ The nth term of AP is an = Sn –Sn–1 ⇒ an = \(\frac{1}2\) (3n2 + 7n) –\(\frac{1}2\) [3(n − 1)2 + 7(n − 1)] = \(\frac{1}2\) [3n2 + 7n – 3(n − 1)2 − 7(n − 1)] = \(\frac{1}2\) [3n2 + 7n – 3(n2 − 2n + 1) − 7(n − 1)] = \(\frac{1}2\) [3n2 + 7n – 3n2 + 6n – 3 – 7n + 7] = \(\frac{1}2\) [6n + 4] = 3n + 2. Hence, the nth terms of given AP is an = 3n + 2. Now, put n = 20, 20th term = 3n + 2 = 3 × 20 + 2 = 60 + 2 = 62. Hence, the 20th term of AP is a20 = 62. |
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