Question

If the sum of p terms of an A.S is q and the sum of q terms is p, show that the sum of (p + q) terms is -(p + q)

Answer 1

Answer:HOPE it helps you ✨✌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?

How good is the noise-cancellation of OPPO Enco X?

One thing we can say for sure – with the all-new OPPO Enco X you hear only what you want to hear. With its powerful Hybrid Active Noise Control (ANC) Technology, you can tune out the world and tune into your favourite music. Let’s talk about its Hybrid ANC technology – this technology uses Feed-Forward (FF) and Feedback (FB) noise control together to ensure maximum noise cancellation width and depth. When using its flagship Max Noise Cancellation Mode, you can enjoy an immersive music experience even in noisy environments, by virtue of its aforementioned Hybrid ANC technology together with excellent passive noise control. Protect your hearing, because even in noisy environments, you don’t have to turn up the volume to focus on the audio. Using the Noise Cancellation Mode, you can enjoy a

Gunjan Gaba

Answered 3 years ago

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). Mark as brainliest.☺✈✨✈✨✈✨✈✨✈

Answer 2

Step-by-step explanation:

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:

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

So, we have

So, we haveS(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). … :-)