Question

show that the sum of (p+q)th and (p-q)th term of an AP is equal to twice its pth term

Answer 1

Let the common difference of the AP = d

Let the pth term = P

=> (p-q)th term = P - d -d - d- d - .............upto q times

=> (p-q)th term = P - dq

Similarly

(p+q)th term = P + d +d + d+ d + .............upto q times

=> (p+q)th term = P + dq

Hence the sum of p+q)th and (p-q)th  term

= P + dq + P - dq

= 2P

Hence the sum of (p+q)th and (p-q)th term of an AP is equal to twice its pth term(proved)

Answer 2

Answer:

Proved

Step-by-step explanation:

show that the sum of (p+q)th and (p-q)th term of an AP is equal to twice its pth term

Let say first term = a

Common difference = d

(p+q) th term = a + (p+q-1)d

(p-q) th term = a + (p=q-1)d

pth term = a + (p-1)d

Sum of (p+q) th term & (p-q) th term

(p+q) th term + (p-q) th term = a + (p+q-1)d + a + (p-q-1)d

= a + pd + qd - d + a + pd -qd -d

= 2a + 2pd - 2d

= 2(a + pd -d)

= 2(a + (p-1)d)

= 2 pth term

QED