Question

Solve plzzz....
100 points...

Questions is in above attached.
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Answer 1

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

TO PROVE :

$\sf{S_{p+q} = \frac{p + q}{2}[ a + b + \frac{a - b}{p - q} ]}$

PROOF :

Let A be the first term and D be the common difference of the A.P.

and It is given that,

$\bf{t_{p} = a}$

$\sf {\implies \: A + (p - 1)D = a} \: \: \rightarrow(1)$

$\bf{t_{q} = b}$

$\sf {\implies \: A + (q- 1)D = b} \: \: \rightarrow(2)$

Subtracting (2) from (1), we get

$\sf {\implies \: \: (p - 1 - q + 1)D = a - b}$

$\sf{ \implies \: D \: = \: \frac{a - b}{p - q}} \: \: \: \rightarrow \: (3)$

Adding (1) and (2), we get

$\sf{ \implies \: 2A + (p + q - 2) D = a + b}$

$\implies \: \sf {\: 2A + (p + q - 1) D = a + b + D}$

Putting the value of D, we get

$\sf{ \implies \: 2A + (p + q - 1) D = a + b + \frac{a - b}{p - q} } \: \: \rightarrow \: (4)$

Now we know ,

$\sf{S_{p+q} = \frac{p + q}{2}[2A + (p + q - 1)D]} \: \: \rightarrow(5)$

From (4) and (5) we get,

$\sf{S_{p+q} = \frac{p + q}{2}[ a + b + \frac{a - b}{p - q} ]}$

HENCE PROVED .

Answer 2

Answer:

HENCE IT IS PROVED

Step-by-step explanation:

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