Sequence And Series Question 385
Question: What is the 15th term of the series 3, 7, 13, 21, 31, 43, ….?
Options:
205
225
238
241
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let $ \begin{aligned} & S=3+7+13+21+31+…..+a_{n} \\ & -S_3=\pm 7\pm 13\pm 21\pm 31\pm …\pm a_{n}\pm a_{n} \\ & 0=3+4+6+8+10+12+….-a_{n} \\ \end{aligned} $
$ \Rightarrow a_{n}=3+[4+6+8+10+12+…(n-1),terms] $ $ =3+\frac{(n-1)}{2}[8+{(n-1)-1}\times 2] $ $ =3+\frac{(n-2)}{2}[8+2n-4] $ $ =3+\frac{(n-2)}{2}[2n+4] $ $ =3+(n-1),(n+2) $
$ \therefore $ 15th term $ =a_{15}=3+(15-1)\cdot(15+2) $ $ =3+14\times 17=241 $
BETA
