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If the sum of n terms of an A.P. is 3n2 + 5n then which of its terms is 164 ?
Answer & Solution
Correct Answer:
Option
B
Sum of n terms of an A.P. = 3n2 + 5n
Let a be the first term and d be the common difference
Sn = 3n2 + 5n
S1 = 3(1)2 + 5 × 1 = 3 + 5 = 8
S2 = 3(2)2 + 5 × 2 = 12 + 10 = 22
∴ First term (a) = 8
a2 = S2 - S1 = 22 - 8 = 14
d = a2 - a1 = 14 - 8 = 6
Now an = a + (n - 1)d
⇒ 164 = 8 + (n - 1) × 6
⇒ 6n - 6 = 164 - 8
⇒ 6n = 156 + 6
⇒ 6n = 162
⇒ n = $$\frac{{162}}{6}$$
⇒ n = 27
∴ 168 is 27th term
Let a be the first term and d be the common difference
Sn = 3n2 + 5n
S1 = 3(1)2 + 5 × 1 = 3 + 5 = 8
S2 = 3(2)2 + 5 × 2 = 12 + 10 = 22
∴ First term (a) = 8
a2 = S2 - S1 = 22 - 8 = 14
d = a2 - a1 = 14 - 8 = 6
Now an = a + (n - 1)d
⇒ 164 = 8 + (n - 1) × 6
⇒ 6n - 6 = 164 - 8
⇒ 6n = 156 + 6
⇒ 6n = 162
⇒ n = $$\frac{{162}}{6}$$
⇒ n = 27
∴ 168 is 27th term
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LoginS1=a1=3(1)2+5(1)=8
S2=3(2)2+5(2)=22
S2=22=a1+a2
a2=22-8=14
d=a2-a1=14-8=6
nth term value is 164,then what is n?
nth term=a+(n-1)d
164=8+(n-1)6
164-8 / 6=n-1
156/6=n-1
26+1=n
n=27
So 27thterm is 164.