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Sum of 'n' Terms When 'a' and 'l' Is Given

Question 10 i...

Question

Question 10 (ii)
Show that $a_{1}, a_{2} \dots \dots, a_{n}, \dots \dots$ form an AP where $a_{n}$ is defined as below.
(ii) $a_{n} = 9 - 5 n$
Also find the sum of the first 15 terms in each case.

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Solution

$a_{n} = 9 - 5 n a_{1} = 9 - 5 \times 1 = 9 - 5 = 4 a_{2} = 9 - 5 \times 2 = 9 - 10 = - 1 a_{3} = 9 - 5 \times 3 = 9 - 15 = - 6 a_{4} = 9 - 5 \times 4 = 9 - 20 = - 11 I t c a n b e o b s e r v e d t h a t, a_{2} - a_{1} = - 1 - 4 = - 5 a_{3} - a_{2} = - 6 - (- 1) = - 5 a_{4} - a_{3} = - 11 - (- 6) = - 5 i . e ., a_{k + 1} - a_{k} i s s a m e e v e r y t i m e . T h e r e f o r e, t h i s i s a n A P w i t h c o m m o n d i f f e r e n c e a s - 5 a n d f i r s t t e r m a s 4. S_{n} = \frac{n}{2} [2 a + (n - 1) d] S_{15} = \frac{15}{2} [2 (4) + (15 - 1) (- 5)] = \frac{15}{2} [8 + 14 (- 5)] = \frac{15}{2} (8 - 70) = \frac{15}{2} (- 62) = 15 (- 31) = - 465$

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

a_{n}

a_{n} = 9 - 5 n

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(i) a_n = 3 + 4n

(ii) a_n = 9 − 5n

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

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