Question

How many term of the A.P. 2,6,10.....are needed to make the sum 288

Answer 1

Heya !!

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A.P. 2, 6, 10, …

a = 2

d = 6-2 = 4

Sn = n/2 [ 2a + (n-1) d ]

288 = n/2 [ 2(2) + (n-1) 4 ]

=> 2×288 = n ( 4 + 4n - 4 )

=> 576 = n (4n)

=> 576 = 4n²

=> n² = 576/4

=> n² = 144

=> n = √144

=> n = 12

Hence, 12 terms are needed to give the sum 288.

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Hope my ans.'s satisfactory.☺

Answer 2

Solution:

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Given:

AP: 2, 6 , 10,...

∴ a =2,

∴ d = 6-2 =4,

&

$S_n = 288$

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To find:

 The value of n,.

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We know that,

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

=> $288 = \frac{n}{2}(2(2)+ (n-1)4)$

=> $288 = \frac{n}{2} (4+4n -4)$

=> $288 = \frac{n}{2} (4n)$

=> $288 = n(2n)$

=> $288 = 2n^2$

=> $n^2 = 144$

As, number of terms can't be negative,.

=> ∴ n = + 12,..
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                             => Hope it Helps !!