Question

in an a.p. 6th term is half the 4th term and 3rd term is 15. how many terms are needed to give a sum of 66.

Answer 1

Heya Frnd..........☺

a6 = ½ a4

= a + 5d = ½ (a + 3d )

= 2a + 10d = a + 3d

=>. a = - 7d ............( 1 )

NOW,

a + 2d = 15

=> -7d + 2d = 15

=> d = 15/-5

=> d = -3 ...............( 2 )

So,

a = 21 ..............( 3 )

NOW,

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

=> 66 = n/2 ( 2(21) + (n-1)(-3))

=> 132 = n [ 42 -3n + 3]

=> 132 = 45n - 3n²

So,

n² - 15n + 44 = 0

=> n² - 11n - 4n + 44

=> n ( n - 11) - 4(n - 11)

=> (n - 11)(n - 4)

So,

n = 11 ............( rejected)

n = 4. ..................★ANS★

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