Question

find the sum of first 11 term of an A.P of which the 6th term is 45

Answer 1

t6 = 45
but, tn = a+(n-1)d
t6.= a +5d
45 = a+5d
a= 45-5d
t11 = a+10d
Sn = n/2 {2a+(n-1)d}
S11 = 11/2 {2×(45-5d) + 10d
S11 = 11/2 {90-10d +10d}
S11 = 11/2 × 90
S11 = 11×45
S11 = 495

Answer 2

The sum of first 11 term of an A.P of which the 6th term is 45 is 495

Step-by-step explanation:

Formula of nth term of AP = $a_n=a+(n-1)d$

where a is the first term

d is the common difference

n = No. of terms

$a_n$ = nth term

Substitute n = 6

$a_6=a+(6-1)d$

$a_6=a+5d$

6th term is 45

So, a+5d=45  --- 1

Formula of sum of first n terms =$S_n=\frac{n}{2}(2a+(n-1)d)$

Substitute n = 11

$S_{11}=\frac{11}{2}(2a+(11-1)d)$

$S_{11}=\frac{11}{2}(2a+10d)$

$S_{11}=\frac{11}{2}(2)(a+5d)$

Using 1

$S_{11}=\frac{11}{2}(2)(45)$

$S_{11}=495$

Hence  the sum of first 11 term of an A.P of which the 6th term is 45 is 495

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