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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# 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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a= 9−5_{n}na_{1}= 9−5×1 = 9−5 = 4a_{2}= 9−5×2 = 9−10 = −1a_{3}= 9−5×3 = 9−15 = −6a_{4}= 9−5×4 = 9−20 = −11We can see here, the common difference between the terms are;

a_{2}−a_{1}= −1−4 = −5a_{3}−a_{2}= −6−(−1) = −5a_{4}−a_{3}= −11−(−6) = −5Hence,

a_{k}_{ + 1}−ais same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4._{k}Now, we know, the sum of nth term is;

S=_{n}n/2 [2a+(n-1)d]S= 15/2[2(4) +(15 -1)(-5)]_{15 }= 15/2[8 +14(-5)]

= 15/2(8-70)

= 15/2(-62)

= 15(-31)

= -465