Question

Sum of first n terms of an A.P is 6n2+6n. Then find 4th term of series. a) 120 b) 72 c) 48 d) 24

Answer 1

Answer:

Step-by-step explanation:

Sn = 6n²+6n

= 6n(n+1)

N¹ = 6×1(1+1)

= 6×2 = 12

N²=6×2(2+1)

= 12×3 = 36

N³=6×3(3+1)

=18×4 = 72

A=12 , a² = 36-12=24 , a³ = 72-36 = 36

A. P formed is 12 , 24 ,36 ...

A4 = a+(n-1)d

= 12+(4-1)12

=12+(3×12)

=12+36 = 48

Hope it works

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Answer 2

Answer:

The 4th term of the series is option(c) 48.

Step-by-step explanation:

Sum of first n terms of an AP = $6n^2+6n$ (As Given in the question)

By substituting n=1, we get the first term of the AP

First term = $6\times1^2+6\times1$

=12

By substituting n=2, we get the sum of first    two terms of the AP

Sum of first two terms= $6\times2^{2}+6\times2$=36

Second term= Sum of first two terms - First term  

= 36-12

= 24

By substituting n=3, we get the sum of first three terms of the AP

Sum of first three terms = $6\times3^{2}+6\times3$= 72

Third term = Sum of first three terms - Sum of first two terms

=72-36

=36

By substituting n=4, we get the sum of the first four terms of the AP

Sum of the first four terms= $6\times4^{2}+6\times4$=120

Fourth term= Sum of first four terms- Sum of first three terms

=120-72

=48

The fourth term of the AP is 48.

The correct answer is option (c) 48.