Question

The 14th term for an a.p. is twice its 8th term. If its 6th term is -8, then find the sum of its first 20 term.

Answer 1

here a14=2a8

A6=-8

s20=?

a =a+(n-1)d

a14=2a8

a+(14-1)d=2{a+(8-1)d}

a+13d=2a+14d

a-2a=14d-13d

-a=d

a=-d

since

A6=a+(6-1)d

-8=-d+5d

-8=4d

d=-2

and

a=-d

a=2

then

s20=n\2{2a+(n-1)d}

s20=20\2{2*2+(20-1)(-2)}

s20=10{4+19(-2)}

s20=10{4-38}

s20=10{-34}

s20=-340

Answer 2

Answer:

-340

Step-by-step explanation:

Given AP, where

• a₁₄=2a₈, a₆= -8, S₂₀=?

Use of formulas:

• aₙ=a+(n-1)d, Sₙ= 1/2n(2a+(n-1)d)

Solution:

• a₁₄= a+13d
• a₈= a+7d
• a+13d=2(a+7d)
• a+13d=2a+14d
• a=-d
• a₆=a+5d= -d+5d=4d
• 4d=-8
• d= -2

• a=-d=2

Required sum of first 20 terms:

• S₂₀=1/2*20*(2a+19d)=
• 10*(2*2-19*2)=
• 10*(-34)= -340

Answer: -340