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If first, second and last terms are respectively a, b and 2a of an arithmetic progression, then prove that sum of progression will be3ab 2(b- a)​

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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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