Question

If the sum of first n terms of an
a.p is given by sn = 3n2 + 2n, find the nth term of the
a.p.

Answer 1

given  sum of n terms = 3n² + 2n

let first term be a₁, second term be a₂ and so on.....

if n = 1

sum = a₁ (since there is no sum when considering only 1st term)

therefore, a₁ = 3(1)² + 2(1) = 3 + 2 = 5

then, if n = 2

S₂ = a₁ + a₂ = 3(2)²  + 2(2) = 12 + 4 = 16

since a₁ = 5

∴ a₂ = 16 - 5 = 11

similarly, if n =  3

S₃ =  3(3)² + 2(3) = 27 + 6 = 33

∴ a₃ = 33 - (a₁ + a₂) = 33 - 16 = 17

∴ we gain that common difference = d = 17 - 11 = 11- 5 = 6

∴ let nth term be an

an = a₁ + (n-1)d

( where a₁ = 5 ; d = 6 and n = n)

∴ an = 5 + (n-1)6

        = 5 + 6n - 6

        = 6n - 1

This is your answer. Hope it helps you.

Answer 2

Answer:

The 15th term is 89 and AP is 5,11,17,23,...

Step-by-step explanation:

Given: The sum of first n terms of an a.p is given by$sn = 3n^2 + 2n$.

We have to determine the A.P. and its 15th term.

• As we know, the nth term of an arithmetic sequence is given by.

           $a_n$= a + (n – 1)d.

• The number d is called the common difference because it is any two consecutive terms of$a_{n}$.

• the arithmetic sequence differs by d, and it is found by subtracting any pair of terms$a_{n}$  and $a_n_+_1$.

We are solving in the following way:

We have,

The sum of first n terms of an a.p is given by$sn = 3n^2 + 2n$.

Taking n=1 we get,

[tex]\begin{array}{l} \mathrm{S}_{1}=3(1)^{2}+2(1) \\ \Rightarrow \mathrm{S}_{1}=3+2 \\ \Rightarrow \mathrm{S}_{1}=5 \\ \Rightarrow \mathrm{a}_{1}=5 \end{array}[/tex]

Taking n=2 we get,

[tex]\begin{array}{l} \mathrm{S}_{2}=3(2)^{2}+2(2) \\ \Rightarrow \mathrm{S}_{2}=12+4 \\ \Rightarrow \mathrm{S}_{2}=16 \\ \therefore \mathrm{a}_{2}=\mathrm{S}_{2}-\mathrm{S}_{1}=16-5=11 \end{array}[/tex]

Taking n=3 we get,

[tex]\begin{array}{l} \mathrm{S}_{3}=3(3)^{2}+2(3) \\ \Rightarrow \mathrm{S}_{3}=27+6 \\ \Rightarrow \mathrm{S}_{3}=33 \\ \therefore \mathrm{a}_{3}=\mathrm{S}_{3}-\mathrm{S}_{2}=33-16=17 \end{array}[/tex]

So, a=5,

$\mathrm{d}=\mathrm{a}_{2}-\mathrm{a}_{1}=11-5=6$

Now, the 15th term will be:

[tex]\begin{array}{l} a_{n}=a+(n-1) d \\ a_{15}=5+(15-1) 6 \\ a_{15}=5+14 \times 6 \\ a_{15}=5+84 \\ a_{15}=89 \end{array}[/tex]

Hence, the 15th term is 89 and AP is 5,11,17,23,...

NOTE:

Q. The sum of the first n term of an A.P. is given by  $sn = 3n^2 + 2n$,

Determine the A.P. and its 15th term.