# pth term of an A.P. is 'a' and qth term is 'b'. prove that sum of (p+q)th term is p+q/2[a+b+(a-b/p-q)]. plz ans needed soon.....

i) Let the first term of the AP be 'c' and its common difference be 'd'.

ii) So its pth term is: c + (p - 1)d = a ------ (1)
and its qth term is: c + (q - 1)d = b ----- (2)

Adding (1) & (2): 2c + (p + q - 2)d = a + b

==> 2c + (p + q - 1)d - d = a + b

==> 2c + (p + q - 1)d = (a + b) + d ---------- (3)

iii) Operating (1) - (2): (p - q)d = (a - b)
==> d = (a - b)/(p - q) --------- (4)

iv) Sum to (p + q) terms is:
S₍p+q₎ = {(p + q)/2}*{2c + (p + q - 1)d}

Substituting from (3) and (4), we get

S₍p+q₎ = {(p + q)/2}*{(a + b) + (a - b)/(p - q)} ---- [Proved]
• 25
Let first term be a and common difference be d.

pth term = a + (p-1)d = q -(i)

qth term = a?+ (q-1)d = p -(ii)

(ii) - (i) gives

(q - p)d = p - q or d = -1

(i) gives? a = q - (p-1)d = q - (p-1)(-1) = q + p - 1

Hence,

nth term = a + (n-1)d = q + p -1 + (n-1)(-1) = q + p -1 + n +1 = p + q - n
• -12
@V. Harshith How did the step..., which is before the equation 3... Came
• 1
Here is solution....

• 14
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