Math, asked by asakeena, 1 year ago

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


MegaRayquaza16: r u retarded or something?
MegaRayquaza16: why r u repeating the question

Answers

Answered by Anonymous
0
Heya !!

==================================

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.

==================================

Hope my ans.'s satisfactory.☺
Answered by sivaprasath
0
Solution:

_____________________________________________________________


Given:

AP: 2, 6 , 10,...

∴ a =2,

∴ d = 6-2 =4,

&

S_n = 288

_____________________________________________________________

To find:

 The value of n,.

_____________________________________________________________


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,..
_____________________________________________________________

                             => Hope it Helps !!

sivaprasath: Mark as Brainliest,.
Similar questions