Question

The nth term of an ap is given by (-4n+15).Find the sum of first 20 terms of this A.P.

Answer 1

Answers:

$\huge{\pink{\boxed{\green{\sf{S_{20}=-540}}}}}$

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

$\: \: \: \: \: \: \: \: \: \: { \orange{given}} \\ { \pink{ \boxed{ \green{ a_{n} = ( - 4n + 15) }}}} \\ \\{ \blue{ to \: find}} \\ { \purple{ \boxed{ \red{S _{20} = ? }}}}$

☆ According to given question:

☆ Let n =1,2,3,4,5,6.....

$\to n = 1\\ \to a_{1 } = - 4 \times 1 + 15 \\ \to a_{1} = - 4 + 15 \\ { \boxed{\to a_{1} = 11 }}\\ \\ \to n = 2 \\ \to a_{2} = - 4 \times 2 + 15 \\ \to a _{2} = - 8 + 15 \\{ \boxed{ \to a _{2} = 7 }}\\ \\ \to n = 3 \\ \to a _{3} = - 4 \times 3 + 15 \\ \to a _{3} = - 12 + 15 \\ { \boxed{\to a _{3} = 3}} \\ \\ \therefore a. p = 11.7.3...... \\ \to common \: difference = a_{2} - a_{1} = 7 - 11 = - 4 \\ \to a1 = 11$

☆ So we find the value of first term, common difference.

☆ We know the formula for sum of nth terms.

$\to S_{n} = \frac{n}{2} (2a + (n - 1)d) \\ \to S_{20} = \frac{20}{2} (2 \times 11 + (20 - 1) \times - 4) \\ \to S_{20} =10(22 + 19 \times - 4) \\ \to S_{20} =10(22 - 76) \\ \to S_{20} =10( - 54) \\ { \pink{ \boxed{ \green{\therefore S_{20} = - 540}}}}$

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