Question

Hey guys ❤
The 14th term of an AP is twice its 8th term.if it's 6th term is -8 then the sum of its 20 terms is ​

Answer 1

Solution:-

Given

$\to \rm \: T_{14} = 2T_8$

$\rm \to \: T_{6} = - 8$

Now we can write as

$\rm \: \to \: a + 13d = 2(a + 7d)$

$\rm \: a + 13d = 2a + 14d$

$\rm \: \to a + d = 0 \: \: \: \: \: \: ...(i)$

$\rm \to \: a + 5d = - 8 \: \: \: \: \: \: ...(ii)$

Now using substitution method

Taking eq ( i )

$\rm \: a = - d \: \: \: ....(iii)$

Now substitute (iii) eq on (ii) eq

$\rm \to \: a + 5d = - 8 \: \: \: \: \: \: ...(ii)$

$\to \rm \: - d + 5d = - 8$

$\rm \to \: 4d \: = \: - 8$

$\boxed{ \rm d \: = - 2}$

$\boxed{ \rm \: a = 2}$

Now we have to find

$\to \: \rm S_{20}$

Formula

$\rm \: S_{n} = \dfrac{n}{2} \{2a + (n - 1)d \}$

$\rm \: S_{20} = \dfrac{20}{2} \{2 \times 2 + (20 - 1) \times - 2 \}$

$= 10 \{4 + 19 \times - 2 \}$

$= 10 \{4 - 38 \}$

$\rm \: = 10 \{ - 34 \}$

$\rm \: = - 340$

$\rm \: S_{20} = - 340$