Question

If the sum of n terms of an

a.p. is 3n2 + 5n and its mth term is 164, find the value of m.

Answer 1

3n^2+5n
put n=1,2,3,4.....
3×1^2+5×1
sn=8
s2=22
s3=52
t1=a=8
t2=22-8
t2=14
t3=42-22
t3=20
d=14-8=6
tm=a+(m-1)d
164=8+(m-1)6
164-8=6m-6
156+6=6m
162=6m
m=27

Answer 2

Answer:

27

Step-by-step explanation:

Concept

• Arithmetic progression is defined as a sequence of numbers, for every pair of consecutive terms, we get the second number by adding a constant to the first one.

Given

$3n^{2} +5n$ and nth term is 164

Find

Value of n

Solution

Let $S_{n}$ denote the sum of n terms and $a_{n}$ be the nth term of the given A P. Then,

$\mathrm{S}_{\mathrm{n}}=3 \mathrm{n}^{2}+5 \mathrm{n}$

$\mathrm{S}_{\mathrm{n}-1}=3(\mathrm{n}-1)^{2}+5(\mathrm{n}-1)=3 \mathrm{n}^{2}-\mathrm{n}-2$

Now,$a_{n}=S_{n}-S_{n-1}$

$a_{n}=\left(3 n^{2}+5 n\right)-\left(3 n^{2}-n-2\right)$

$a_{n}=6 n+2$

Now, $a_{m}=164$

$6 m+2=164$

$6 m=162$

$m=27$

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