Question

Find the sum of 51 terms of an ap in which d=7and 22nd term is 149

Answer 1

↑ Here is your answer ↓
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$a_{22} = 149$

$a+21d = 149$

Given that d = 7

$a+21(7) = 149$

$a+147 = 149$

$a=2$

Now, sum of first 51 terms

$S_{51} = \dfrac {51} {2} [2(2)+(51 - 1)7]$

$S_{51} = \dfrac {51}{2} [4 + 50(7)]$

$S_{51} = \dfrac {51}{2} [4+350]$

$S_{51} = \dfrac {51}{2} * 354$

$S_{51} = 9027$

Answer 2

Hey mate !!

Here's the answer !!

Given :

Common Difference ( d ) = 7

The 22 nd term ( a₂₂ ) = 149

To find : 

Sum of 51 terms of that AP ( S₅₁ = ? )

Proof : 

a₂₂ = a + 21d

We know that d = 7. Hence substituting in the above equation, we get,

149 = a + 21 × 7

149 = a + 147

149 - 147 = a

=> a = 2

Hence the First term of the AP is 2 and the common difference is 7.

Hence we can easily find the sum of 51 terms by using the following formula.

[tex] S_{n} = \frac{n}{2} [ 2a + ( n - 1 ) d ] [/tex]

So here we know that, 

n = 51, a = 2, d = 7

Hence substituting them in the formula we get,

S₅₁ = 51 / 2 [ 2 × 2 + ( 51 - 1 ) × 7 ]

S₅₁ = 51 / 2 [ 4 + ( 50 × 7 ) ]

S₅₁ = 51 / 2 [ 4 + 350 ]

S₅₁ = 51 / 2 × 354

S₅₁ = 25.5 × 354

S₅₁ = 9027

Hence the sum of 51 terms of that AP is 9027.

Hope my answer helps !!

Cheers !!