Question

Find the value of x if 12th term of the A.P.: x-7,x-2,x+3,........ Is 81. Find S12 also.

Answer 1

t12 = a+11d
81 =x-7 +11×5
81=x-7 +55
81-55=x-7
26=x-7
x=26+7
x=33
26 , 31, 36 ,...........
s12= 12/2 [ 2 (26)+11 (5)]
s12=6 [52+55]
s12= 6 (107)
s12=642

Answer 2

Concept

An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

$a_{n}=a_{1}+(n-1) d$

$a_{n}=$the $n^{\text {th }}$term in the sequence

$a_{1}=$the first term in the sequence

$d=$ the common difference between terms

Given

The series  x-7,x-2,x+3,........ and 12th term is 81.

Find

We have to find the value of x and find the value is $S_{12}$.

Solution

First find the value of d

d= (x-2)-(x-7)

 = 5

We know  12th term is 81

then,

81 =  a+11d

 81 =x-7 +11×5

81=x-7 +55

81-55=x-7

26=x-7

x=26+7

x=33

then a= x-7

         =33-7

         =26

And the sum of AP is S = n/2[2a + (n − 1) × d]

find the value of $S_{12}$

$s_{12}= 12/2 [ 2 (26)+11 (5)]$

$s_{12}=6 [52+55]$

$s_{12}= 6 (107)$

$s_{12}=642$

Hence the value of x is 33 and $S_{12}$ is 642.