Question

in an ap the 4th and the 6th term are 8and 14find the sum of first 20term​

Answer 1

GIVEN :–

• 4th term of A.P. = 8

• 6th term of A.P. = 14

TO FIND :–

• Sum of first 20term = ?

SOLUTION :–

• If 'a' is first term and 'd' is Common difference , then nth term –

$\\ \longrightarrow \: \large{ \boxed{ \sf T_{n} = a + (n - 1)d}}$

• According to the first condition –

$\\ \implies \sf T_{4} = 8$

$\\ \implies \sf a + (4 - 1)d = 8$

$\\ \implies \sf a + 3d = 8$

$\\ \implies \sf a= 8 - 3d \: \: - - - eq.(1)$

• According to the second condition –

$\\ \implies \sf T_{6} = 14$

$\\ \implies \sf a + (6 - 1)d = 14$

$\\ \implies \sf a + 5d = 14$

• Using eq.(1) –

$\\ \implies \sf (8 - 3d) + 5d = 14$

$\\ \implies \sf 8 + 2d = 14$

$\\ \implies \sf 2d = 14 - 8$

$\\ \implies \sf 2d = 6$

$\\ \: \: \to \: \: \large{ \boxed{ \sf d = 3}}$

• Put the value of 'd' in eq.(1) –

$\\ \implies \sf a= 8 - 3(3) \:$

$\\ \implies \sf a= 8 - 9 \:$

$\\ \: \: \to \: \: \large{ \boxed{ \sf a = - 1}}$

• We know that sum of n terms –

$\\ \longrightarrow \: \large{ \boxed{ \sf S_{n} = \dfrac{n}{2} [2a + (n - 1)d]}}$

• So that , Sum of first 20 terms –

$\\ \implies \sf S_{20} = \dfrac{20}{2} [2( - 1) + (20- 1)(3)]$

$\\ \implies \sf S_{20} = (10)[ - 2 + (19)(3)]$

$\\ \implies \sf S_{20} = (10)[ - 2 + 57]$

$\\ \implies \sf S_{20} = (10)(55)$

$\\ \longrightarrow \: \large{ \boxed{ \sf S_{20} =550}}$

Answer 2

${\mathcal{ANSWER:-}}$

${\color{grey}{\sf{Given:-}}}$

a₄ = 8 and a₆ = 14

${\color{grey}{\sf{To~find:-}}}$

The sum of first 20 terms (S₂₀).

${\color{grey}{\sf{Solution:-}}}$

We know that,

★ a₄ = a + 3d = 8⠀⠀⠀⠀⠀.............(i)

★ a₆ = a + 5d = 14⠀⠀⠀⠀.............(ii)

Subtracting (i) from (ii), we get:-

⇝2d = 6

⇝d = 6/2 = 3

${\therefore}~{\color{magenta}{\sf{Common~difference (d) = 3}}}$

Substituting d = 3 in (i):-

⇝a + 3 × 3 = 8

⇝a + 9 = 8

⇝a = 8 - 9

⇝a = -1

${\therefore}~{\color{magenta}{\sf{First~term (a) = -1}}}$

We know that,

★ Sn = n/2{2a + (n-1)d}

${\sf{Here~n = 20}}$

⇝S₂₀ = 20/2 {2×(-1) + (20-1)3}

⇝S₂₀ = 10 (-2 + 19 × 3)

⇝S₂₀ = 10 (-2 + 57)

⇝S₂₀ = 10 × 55

⇝S₂₀ = 550

${\therefore}~ \red{\sf{Sum~of~first~20 ~terms=550}}$