Math, asked by sonkarg290, 5 months ago

If first, second and last terms are respectively a, b and 2a of an arithmetic progression, then prove that sum of progression will be
3ab 2(b- a)





Answers

Answered by Yugant1913
12

Answer:

Answer:Correct question

If first, second and last terms are respectively a, b and 2a of an arithmetic progression, then prove that sum of progression will be

3ab/ 2(b- a)

Step-by-step explanation:

First term = a, second term = b

Last term = 2a

∴common difference d = b - a

∴ From \:  \:  \:  \:  \:  \:  a_{n}  = a + (n - 1)d \\

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: 2a = a + (n - 1)(b - a)

⇒ \:  \:  \:  \:  \: 2a - a = (n - 1)(b - a)

⇒ \:  \:  \:  \:  \:  \:  \:  \:  \:  \: a = (n - 1)(b - a)

⇒ \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: n - 1 =  \frac{a }{b - a}  \\

⇒ \:  \:  \:  \:  \:  \:  \: n = 1 +  \frac{a}{b - a}  \\  \\  =  \frac{b - a + a}{b - a}  =  \frac{b}{b - a}

∴ From, \:  \:  \:  \:  S _{n}  =  \frac{n}{2} [a + l]  \\

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \frac{b}{2(b - a)}  \times (a + 2a) \\

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \frac{b \times 3a}{2(b - a)}  \\

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \frac{3ab}{2(b - a)}  \\

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \huge\sf\mathbb\color{black} \underline{\colorbox{red}{☠proved☠}}

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