Question

if the 5th and 12th term of an ap are - 4 and -18 respectively find the sum of first 20 term of the AP

Answer 1

t5=-4

t12=-18

so,

a+4d=-4.........1)

a+11d=-18.......2)

on subtracting 1 from 2 we have:

7d=-14

d=-2

on putting value of d in eq 1 we have:

a=4

now,

S20=20/2{2a+(20-1)d}

=10{8-38}

=10{-30}

=-300

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Answer 2

Thanks for asking question


According to the question we know that the 5th and 12th term of an AP is -4,-18

So, t5 = -4

=> We can say that a + 4d = -4 .....(I)

and, t12= -18 Then, a+11d = -18.....(II)

Now, we can subtracted (II) -(I) Then,

d = -2

Now, we can put it value in given equation

a + 4d => a + 4×-2 => a = 4

Then, we know that formula for sums of n terms = Sn = n/2[2a+(n-1)d]

So,

$sn = \frac{20}{2}(2 \times 4(20 - 1)( - 2)$


$= 10 (8 + (20 - 1) \times ( -2 )$

$= 10(8 - 38)$


$= 80 - 380 = - 300$

I hope it's help you