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[Solved] Show that a1, a2 …, an, … form an AP where an is defined as below (i) an = 3+4n (ii) an = 9−5n

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Show that a1, a…, an, … form an AP where an is defined as below

(i) an = 3+4n

(ii) an = 9−5n

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

1 Answer
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(i) an = 3+4n

a1 = 3+4(1) = 7

a2 = 3+4(2) = 3+8 = 11

a3 = 3+4(3) = 3+12 = 15

a4 = 3+4(4) = 3+16 = 19

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

a2 − a1 = 11−7 = 4

a3 − a2 = 15−11 = 4

a4 − a3 = 19−15 = 4

Hence, ak + 1 − ak is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.

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

Sn = n/2[2a+(n -1)d]

S15 = 15/2[2(7)+(15-1)×4]

= 15/2[(14)+56]

= 15/2(70)

= 15×35

= 525

(ii) 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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