Question

If the sum of n terms of an AP is (3n² + 5n) and its m^th term is 164, find the value of m.

Answer 1

To Find: m

Given: Sum of n terms, m^th term

Put n = 1 to get the first term

So a₁ = 3 + 5 = 8

Put n = 2 to get the sum of first and second term

So a₁ + a₂ = 12 + 10 = 22

So a₂ = 14

Common difference = 14 - 8 = 6

T_n = a + (n - 1)d = 8 + (n - 1)6 = 6n + 2

Now 6m + 2 = 164

Or m = 27

The value of m is 27.

Answer 2

Let sum of n terms be S_n

Let m^th term be a_m 

We have, 

S_n= 3n²+5n and a_m=164

Put n=1 ,

We get,

         S₁   =  3(1)² + 5(1) = 3+5 = 8      .............1

Put n=2 ,

S₂= a₁+a₂ 

12+10= 8 + a₂ 

.: a_{2 = 14}         

So, d = 6                ....................2

Put 1&2 in a_m 

We get,

164= 8 +(m-1)6

156/6 = m-1

m-1 = 26 

.°. m = 27 

Answer 3

Let sum of n terms be S_n

Let m^th term be a_m 

We have, 

S_n= 3n²+5n and a_m=164

Put n=1 ,

We get,

         S₁   =  3(1)² + 5(1) = 3+5 = 8      .............1

Put n=2 ,

S₂= a₁+a₂ 

12+10= 8 + a₂ 

.: a_{2 = 14}         

So, d = 6                ....................2

Put 1&2 in a_m 

We get,

164= 8 +(m-1)6

156/6 = m-1

m-1 = 26 

.°. m = 27