Question

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❐ If the sum of 'p' terms of an A.P. is 'q' and the sum of 'q' terms is 'p', show that:–
☆ Sum of (p + q) terms is –(p + q)
☆ Sum of (p –q) terms is (p–q)$\sf{ 1 + \dfrac{2q}{p}}$

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Answer 1

Question :

If the sum of 'p' terms of an A.P. is 'q' and the sum of 'q' terms is 'p', show that:–

☆ Sum of (p + q) terms is –(p + q)

☆ Sum of (p –q) terms is (p–q)(1 + 2q/p)

SoIution :

FormuIa used ; Sum of 'n' terms of an A.P. :

$\longmapsto\bold{S_n = \dfrac{n}{2} \ [2a + (n - 1)d]}$

Given sum of 'p' terms of an A.P. is 'q' and sum of 'q' terms is 'p' . So,

$\longmapsto\sf S_p = q = \dfrac{p}{2} \ [\ 2a \ + \ (p - 1)d \ ] \ \ ............ (i) \\\\ \longmapsto\sf S_q = p = \dfrac{q}{2} \ [\ 2a \ + (q - 1)d \ ] \ \ ............(ii)$

And sum of (p + q) terms :

$\longmapsto\sf S_{(p + q)} = \dfrac{p + q}{2} \ [\ 2a + (p + q - 1)d \ ] \ \ ............. (iii)$

From (i) :-

$\longmapsto\sf 2a + (p - 1)d = \dfrac{2q}{p} ............ (iv)$

From (ii) :-

$\longmapsto\sf 2a + (q - 1)d = \dfrac{2p}{p} \ \ ................ (v)$

Now substracting (iv) from (iii) :-

$\longmapsto\sf 2a + (p - q)d - 2a - (q - 1)d = \dfrac{2q}{p} - \dfrac{2p}{q} \\\\\\ \longmapsto\sf d(p - 1 - q + 1) = \dfrac{2q^2 - 2p^2}{pq} \\\\\\ \longmapsto\sf d(p - q) = \dfrac{2(q^2 - p^2)}{pq} \\\\\\ \longmapsto\sf -d(q - p) = \dfrac{2(q + p)(q - p)}{pq} \\\\\\ \longmapsto\sf -d = \dfrac{2(q + p)(q - p)}{pq(q - p)} \\\\\\ \longmapsto\bold{d = \dfrac{-2(q + p)}{pq}} \qquad [\sf MuItipIying \ by \ (-1)]$

Now putting vaIues of 'd' in (iii) :-

$\longmapsto\sf S_{(p + q)} = \dfrac{p + q}{2} \ [2a \ + \ (p - 1)d + qd] \\\\\\ \longmapsto\sf S_{(p + q)} = \dfrac{p + q}{2} \ \bigg[\dfrac{2q}{p} \ + q \bigg(\dfrac{-2(q + p)}{pq} \bigg) \bigg] \\\\\\ \longmapsto\sf S_{(p + q)} = \dfrac{p + q}{2} \ \bigg[\dfrac{2q - 2q - 2p}{p} \bigg] \\\\\\ \longmapsto\sf S_{(p + q)} = \dfrac{p + q}{2} \ \bigg[\dfrac{-2p}{p} \bigg] \\\\\\ \longmapsto\underline{\underline{\boxed{\sf{S_{(p + q)} = -(p + q) }}}} \qquad\qquad [\sf \ Showed \ ]$

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Now sum of (p - q) terms :

Now equation (iii) be Iike :

$\longmapsto\sf S_{(p - q)} = \dfrac{p - q}{2} \ [\ 2a + (p - q - 1)d \ ] \ \ ......... (iii)$

Now we have vaIue of 'd' . Putting it in (iii) :-

$\longmapsto\sf S_{(p - q)} = \dfrac{p - q}{2} \ [2a + (p - 1)d - qd] \\\\\\ \longmapsto\sf S_{(p - q)} = \dfrac{p - q}{2} \ \bigg[ \dfrac{2q}{p} - q \bigg(\dfrac{-2(q + p)}{pq} \bigg) \bigg] \\\\\\ \longmapsto\sf S_{(p - q)} = \dfrac{p - q}{2} \ \bigg[\dfrac{2q + 2q + 2p}{p} \bigg] \\\\\\ \longmapsto\sf S_{(p - q)} = \dfrac{p - q}{2} \ \bigg[\dfrac{2(2q + p)}{p} \bigg] \\\\\\ \longmapsto\sf S_{(p - q)} = (p - q) \ \bigg(\dfrac{2q + p}{p} \bigg)$

$\longmapsto\underline{\underline{\boxed{\sf{S_{(p - q)} = (p - q) \ \bigg(1 + \dfrac{2q}{p} \bigg) }}}} \qquad\qquad [\sf \ Showed \ ]$

Answer 2

$\huge{ \mathfrak{ \overline{ \underline{ \underline{ \red{ Answer☻}}}}}}$

If the sum of p terms of AP is q and the sum of q terms is p, what will the sum of p+q terms be?

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Let the first term of the given AP be ‘a' and the common difference be ‘d'. Then, the sum of first ‘n' terms of the AP is given by:

S(n)= n/2 {2a+(n-1)d} …….(1)

Here, it is given that:

S(p)=q and S(q)=p

Using (1), we get:-

q=p/2 {2a+(p-1)d}

and p= q/2 {2a+(q-1)d}

i.e. 2a+(p-1)d = 2q/p …..(2)

and 2a+(q-1)d = 2p/q …..(3)

Subtracting (3) from (2), we get:

(p-1-q+1)d= 2q/p - 2p/q

So, d= 2(q^2-p^2)/pq(p-q)

i.e. d= -2(p+q)/pq

Now, substituting the value of ‘d' in eq.n (2), we get:

2a + (p-1){-2(p+q)/pq} = 2q/p

i.e. 2a= 2q/p + 2(p-1)(p+q)/pq

This gives:

a= (p^2+q^2-p-q+pq)/pq

So, we have

S(p+q)= (p+q)/2 { 2(p^2+q^2-p-q+pq)/pq - (p+q-1) 2 (p+q)/pq}

i.e. S(p+q)= (p+q)/pq { p^2+q^2-p-q+pq-p^2-pq-qp-q^2+p+q}

So, S(p+q) = -(p+q). … :-)

THANKS FOR ASKING SUCH A QUESTION DEAR FRIEND ☺️✌️