Question

12. The 14" term of an AP is twice the 8" term. If the 6" term is -8, then find the sum of its first 20 terms​

Answer 1

Let ,

• the first term of the A.P be a.
• Let the common difference be d.

We know that the n th term of an A.P is a + (n-1) d

• 14 th term= a + (14-1) d

= a + 13 d

• 8 th term = a + (8-1) d

= a + 7 d

Given 14 th term = 2 × 8 th term

$: \implies \tt \: a+ 13d = 2(a + 7d)$

$: \implies \tt \: a + 13 \: d = 2a + 14 \: d$

$: \implies \tt \: a-2 \: a \: + 13 \: d - 14 d = 0$

$: \implies \tt \: - a - d = 0$

$: \implies \tt \: a+d=0 \: \: \: \: \: \: \: \: \: \: \: \: - (1)$

Also it is given that 6 th term is - 8

$: \implies \tt \: a + (6 - 1) \times d = - 8$

$: \implies \tt \: a + 5d = - 8 \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: - (2)$

Subtracting (2) from (1) we get :

$: \implies \tt \: d - 5d = 0 - (- 8)$

$: \implies \tt \: - 4 \: d = 8$

$: \implies \tt \: d = \frac{8}{ - 4}$

$: \implies \tt \: d = - 2$

Also a + d = 0

$: \implies \tt \: a + (- 2) = O$

$: \implies \tt \: m \: a - 2 = 0$

$: \implies \tt \: a = 2$

Hence we can find the sum of 20 terms easily.

sum of n terms = n/2 x [2 a + (n-1)d]

Put n = 20:

$: \implies\tt \: \frac{20}{2} \: x [2 a + (20-1)d]$

$: \implies\tt \: 10 x [2 × 2 + 19 × (-2)]$

$: \implies\tt \: 10 × [4-38]$

$: \implies\tt \: 10 × [-34]$

$: \implies\tt \: -340$

Answer 2

Complete step-by-step answer:

Note: One may calculate a20 and use the formulaS=n2[a+l]

Where, a1 is the first term, d is the common difference and an is the nth term of an A.P.Form two equations with the help of given two conditions in the problem. Use identity of sum of n terms to get the required answer, given

⇒Sn=n2[2a1+(n−1)d]

n → number of terms

d → common difference

a1→ first term

As we know Arithmetic progression is a sequence of numbers with same successive difference and general term or nth term of A.P. is given

⇒an=a1+(n−1)

Where, an is the nth term

a1 = First term of A.P.

d = common difference

As, we also know that sum of the terms up to n terms of an A.P. is also given

⇒Sn=n2[2a1+(n−1)

Sn = Sum up to n terms.

a1 = First term

d = common difference

Now, coming to the question, we are given that the 14th term of an A.P. is twice of the 8th term and 6th term of the same A.P. is -8 and hence, we have to determine the sum of the A.P. up to 20 terms.

So, we can write equations

⇒a14=2a8

⇒a6=−8

Now, let us suppose the first term of the given A.P. is a1 and the common difference is ‘d’. So, using equation (i) and (iii), we get

⇒a1+(14−1)d=2(a1+(8−1)d)

⇒a1+13d=2(a1+7d)

⇒a1+13d=2a1+14d

⇒13d−14d=2a1−a1=a1⇒a1=−d

And using equation (iv) and (i), we get

⇒a1+(6−1)d=−8

⇒a1+5d=−8

Substituting a1=−d from equation (v) to the equation (vi), we get

−d+5d=−8

⇒4d=−8

⇒d=−2

And, hence value of a1 is given as

a1=2

Now, as we have to find sum of A.P. up to 20 terms, so using equation (ii), we get

S20=202[2×2+(20−1)(−2)]

⇒S20=10[4+19×(−2)]

⇒S20=10[4−38]

⇒S20=10×(−34)

⇒S20=−340

Hence, the sum of given A.P. up to 20 terms is -340.