Show that $a_{1}, a_{2} \dots \dots, a_{n}, \dots \dots$ form an AP where $a_{n}$ is defined as below:
$a_{n} = 3 + 4 n$
Also find the sum of the first 15 terms.

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Solution

$G i v e n, a_{n} = 3 + 4 n a_{1} = 3 + 4 (1) = 7 a_{2} = 3 + 4 (2) = 3 + 8 = 11 a_{3} = 3 + 4 (3) = 3 + 12 = 15 a_{4} = 3 + 4 (4) = 3 + 16 = 19 I t c a n b e o b s e r v e d t h a t; a_{2} - a_{1} = 11 - 7 = 4 a_{3} - a_{2} = 15 - 11 = 4 a_{4} - a_{3} = 19 - 15 = 4 i . e ., a_{k + 1} - a_{k} i s s a m e e v e r y t i m e . T h e r e f o r e, t h i s i s a n A P w i t h c o m m o n d i f f e r e n c e a s 4 a n d f i r s t t e r m a s 7. S_{n} = \frac{n}{2} [2 a + (n - 1) d] S_{15} = \frac{15}{2} [2 (7) + (15 - 1) \times 4] = \frac{15}{2} [(14) + 56] = \frac{15}{2} (70) = 15 \times 35 = 525$

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Similar questions

a_{1}, a_{2} \dots \dots, a_{n}, \dots \dots

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a_{n} = 3 + 4 n

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Also find the sum of the first 15 terms in each case.

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