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Show that a1, a2 … , an , … form an AP where an is defined as below(ii) an = 9−5n Also find the sum of the first 15 terms in each case. Q.10(2)

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How i solve the best way to solve the problem of arithmetic progressions of exercise 5.3 of class 10th , It is very important for class 10th Show that a1, a2 … , an , … form an AP where an is defined as below(ii) an = 9−5n Also find the sum of the first 15 terms in each case.

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  1. an = 9−5n

    a1 = 9−5×1 = 9−5 = 4

    a2 = 9−5×2 = 9−10 = −1

    a3 = 9−5×3 = 9−15 = −6

    a4 = 9−5×4 = 9−20 = −11

    We can see here, the common difference between the terms are;

    a2 − a1 = −1−4 = −5

    a3 − a2 = −6−(−1) = −5

    a4 − a3 = −11−(−6) = −5

    Hence, ak + 1 − ak is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

    Now, we know, the sum of nth term is;

    Sn = n/2 [2a +(n-1)d]

    S15 = 15/2[2(4) +(15 -1)(-5)]

    = 15/2[8 +14(-5)]

    = 15/2(8-70)

    = 15/2(-62)

    = 15(-31)

    = -465

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