Question

If the sum of first n terms of an AP is 4n – n

2 ? Find first term and common difference and 13

th

term?​

Answer 1

Answer:

As it is 4n-n

USING HIT AND TRIAL METHOD

Step-by-step explanation:

PUTTING THE VALUE OF N

1,2,3,4

A1=4*1-1=4-1=3

A2=4*2-1=8-1=7

A3=4*3-1=12-1=11

A4=4*4-1=16-1=15

HERE,

A2-A1=7-3=4

A3-A2=11-7=4

A4-A3=15-11=4

SO,

ALL OF THEM HAS A DIFFERENCE OF 4

SO COMMON DIFFERENCE (D)=4

AND FIRST TERM(a)=3

TO FIND:-

A13=a+(n-1)d

A13=3+(13-1)4

A13=3+12*4

A13=3+48

A13=51

so the 13th term is 51

PLEASE MARK IT AS BRAINLIEST

Answer 2

Answer:

Sum of n terms of AP = 4n - n^2

$\frac{n}{2} (2a + (n - 1)d) = 4n - {n}^{2}$

$2a + (n - 1)d = 4 - 2n$

From the sum

The first term as n=1

a = 4-1 = 3

The sum of two term n=2,

= 8-4 = 4

The sum of three terms n=3,

= 12-9 = 3

So,

a1 = 3

a2 = 4 - 3 = 1

a3 = 3-1-3 = - 1

So the common difference is,

d = a2- a1 = 1 - 3 = - 2

The 13 th term,

$a13 \: = a \: + \: 12d$

$a13 = 3 + 12( - 2) \: = - 21$