Pawan Prajapati
(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
