Question

The sum of first n terms of an AP is 3n2 + 5n. If the mth term is 168 , find the value of m.

Answer 1

S = 3n2 + 5n S1 = a1 = 3+5 = 8 S2 = a1 + a2=6+10 =16                    ⇒a2 = 8 S3 = a1 + a2+ a3  = 9 + 15 = 24        ⇒a3 = 8 ∴ AP is 8,8,8,..... Thus a =8 , d=0           Sm = 164Sum of AP = n/2 [2a + (n-1)d] 164 = m/2 [2(8) + (m-1)0] 164 = m/2[16+0] 164 = m * 16 / 2 m = 164 * 2 /16 m = 20.5

Answer 2

Answer:

Sum of first n terms=

=5n²+3n

Putting n=1, S₁=t₁=5+3=8

Putting n=2, S₂=52²+32=20+6=26

∴, t₂=S₂-S₁=26-8=18

Putting n=3, S³=53²+33=45+9=54

∴, t₃=S₃-S₂=54-26=28

∴, First term=a=8,

common difference=28-18=18-8=10

∴, the mth term=

=8+(m-1)10=168

or, (m-1)10=168-8

or, m-1=160/10

or, m-1=16

or, m=16+1

or, m=17

Step-by-step explanation:

Hope this will help you

please follow me and mark my answer as brainliest answer...