Question

The sum of first n terms of an AP is given by S. = 2n² + 3n. Find the sixteenth term
the AP.​

Answer 1

Question:

The sum of first n terms of an AP is given by S(n) = 2n^2 + 3n. Find the 16th term of the AP.

Answer:

16th term, T(16) = 65

Note:

• An AP is a type of sequence in which the difference between two consecutive terms are equal.

• The nth term of an AP is given by ;

T(n) = a + (n-1)d

where, "a" is the first term and "d" is the common difference of the AP.

• Also, the common difference of an AP is given by ; d = T(n) - T(n-1)

• Also ,the nth term of an AP is given by;

T(n) = S(n) - S(n-1)

Given:

S(n) = 2n^2 + 3n

To find:

16th term ,T(16) = ?

Solution:

It is given that;

S(n) = 2n^2 + 3n

Thus,

=> S(16) = 2(16)^2 + 3(16)

Also,

=> S(15) = 2(15)^2 + 3(15)

Now,

We know that,

The nth term of an AP is given by;

T(n) = S(n) - S(n-1)

Thus,

The 16th term of the AP will be given by;

=> T(16) = S(16) - S(16-1)

=> T(16) = S(16) - S(15)

=> T(16) = [2(16)^2 + 3(16)]

- [2(15)^2 + 3(15)]

=> T(16) = 2(16)^2 + 3(16)

- 2(15)^2 - 3(15)

=> T(16) = 2[(16)^2 - (15)^2] + 3(16 -15)

=> T(16) = 2[(16 +15)(16 -15)] + 3•1

=> T(16) = 2•31•1 + 3

=> T(16) = 62 + 3

=> T(16) = 65

Hence,

The 16th term of the AP is 65 .

Answer 2

Answer:

The sixteenth term of the AP is 65

soln refers to the attachment