Question

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

Answer 1

Answer:

This is the solution ...

Answer 2

$\large\bf\underline{Given:-}$

• 14th term is twice it's 8th term
• 6th term = -8

$\large\bf\underline {To \: find:-}$

• Sum of first 20 terms

$\huge\bf\underline{Solution:-}$

• 6th term = -8
• 14th term = 2(8th term)
• 8th term = a + 7d

14th term = 2( a + 7d)

14th term = 2a + 14d

»» a + 13d = 2a + 14 d

»» a + d = 0 ....(i)

• 6th term = a + 5d
• a + 5d = -8....(ii)

From (i) and (ii)

»» a + d = 0

»» a + 5d = -8

⠀--⠀⠀ --⠀ +

━━━━━━━━━

»» ⠀⠀ - 4d = 8

• »»⠀d = -2

Substituting value of d = -2 in eq (i)

»» a + d = 0

»» a -2 = 0

• »» a = 2

Now,

$\large \bigstar \bf \: S_n = \frac{n}{2} \big\{2a + (n - 1)d \}$

✝️ Sum of first 20 terms :-

$\tt \rightarrowtail \: S_{20} = \frac{20}{2} \bigg\{2 \times 2 + (20 - 1) \times ( - 2) \bigg \} \\ \\ \tt \rightarrowtail \: S_{20} =10 \bigg \{ 4 + (19) \times ( - 2)\bigg \} \\ \\ \tt \rightarrowtail \: S_{20} =10 \bigg \{4 - 38 \bigg \} \\ \\ \tt \rightarrowtail \: S_{20} =10 \times ( - 34) \\ \\ \tt \rightarrowtail \: S_{20} = - 340$

So,

✝️Sum of first 20 terms is - 340.