Question

) Sum of first 16 terms of the AP 10, 6, 2, ......... is ________ *​

Answer 1

$d = 4 \\ a= 10 \\ s = \frac{n}{2}(2a + (n - 1)d) \\ s = \frac{16}{2} (2 \times 10 + (16 - 1)4) \\ s= 8(20 + 60) \\ = 8 \times 80 = 640$

Answer is 640

Answer 2

Given :

AP series 10,6,2..........

Find out :

sum of first 16 terms of AP .

Formula used :

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

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Solution :

a = 10

d = a2 - a1

d = 6 - 10

d = -4

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n = 16

S16 = n/2( 2a + (n -1) d )

S16 = 16/2{ 2(10) + (16-1) -4 }

S16 = 8 { 20 + (15) -4 }

S16 = 8 { 20 - 60 }

S16 = 8 { -40 }

S16 = - 320

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Extra information :

Last term ( an ) = a + ( n -1 ) d

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