Math, asked by sujalrajkbk, 8 months ago

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

Answers

Answered by BrainlyPopularman
54

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}}\\

Answered by Anonymous
26

{\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}}

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