Question

The sum of nth terms of a progression is 3n^2+4n. Is this progression an A.P.? if so, find the A.P. and the sum of its rth term​

Answer 1

Answer:

sum of its rth term​ = 3r² + 4r

Step-by-step explanation:

Sn = 3n²+4n (given)

For n=1

S1 = a1 = 3(1)² + 4(1) = 3+4 = 7

For n=2

S2 = 3(2)² + 4(2) = 12 + 8 =  20

a2 = S2 - S1 = 20 - 7 = 13

For n=3

S3 = 3(3)² + 4(3) = 27 + 12 = 39

a3 = S3 - S2 = 39 - 20 = 19

Sequence is 7, 13, 19...

Now d = a2 - a1 = 13-7 = 6

And d = a3 - a2 = 19-13 = 6

As the difference is same, this progression is an A.P.

We can check for n=4, (just to confirm)

S4 = 3(4)² + 4(4) = 48 + 16 = 64

a4 = S4 - S3 = 64 - 39 = 25

And 25-19= 6 = d

Hence our AP is 7, 13, 19, 25.....

∴ a = 7, d= 6

∴sum of its rth term​ = r[2a+(r-1)d]/2

= r[2×7 +(r-1)6]/2

= (r/2)[14 + 6r -6] = (r/2)(8+6r)

=r(4+3r)

= 3r² + 4r

Note:

This could have been easily obtained by replacing n with r in the given sum

Answer 2

Step-by-step explanation:

sn = 3(n-1)²+4(n-1)

= 3n²-2n-1

tn = (3n²-4n) - ( 3n²- 2n - 1 )

= 6n+1

T1 = 7

T2 = 13

T3 = 19

$a = 7 \: \: d = 6$