24. If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
24. If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.
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1 Answer
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24. Let a and d be the first elements and common different of the A.P.
Then, Sum of first p terms of the AP=Sum of first q terms of A.P
Sp=Sq
p [2a+pd – d]=q [2a+qd – d]
2ap+p2d – pd=2aq+q2d – qd
2ap – 2aq+p2d – q2d – pd+qd=0
2a (p – q)+ [p2 – q2 – p+q]d=0.
2a (p – q)+ [ (p – q) (p+q) – (p – q)]d=0
(p – q) {2a+ [ (p+q) – 1]d}=0
Deviding both sides by p – q,
2a+ [ (p+q) –1]d=0.
And multiplying by P+Q/2 we get,
which in the form where n=p+q.
Sp+q=0.
Similar Questions for you
First term = a
Common difference = d
Given: a + 5d = 2 . (1)
Product (P) = (a1a5a4) = a (a + 4d) (a + 3d)
Using (1)
P = (2 – 5d) (2 – d) (2 – 2d)
-> = (2 – 5d) (2 –d) (– 2) + (2 – 5d) (2 – 2d) (– 1) + (– 5) (2 – d) (2 – 2d)
= –2 [ (d – 2) (5d – 2) + (d – 1) (5d – 2) + (d – 1) (5d – 2) + 5 (d – 1) (d – 2)]
= –2 [15d2 – 34d + 16]
at
-> d = 1.6
a, ar, ar2, ….ar63
a+ar+ar2 +….+ar63 = 7 [a + ar2 + ar4 +.+ar62]
1 + r = 7
r = 6
S20 = [2a + 19d] = 790
2a + 19d = 79 . (1)
2a + 9d = 29 . (2)
from (1) and (2) a = –8, d = 5
= 405 – 10
= 395
3, 7, 11, 15, 19, 23, 27, . 403 = AP1
2, 5, 8, 11, 14, 17, 20, 23, . 401 = AP2
so common terms A.P.
11, 23, 35, ., 395
->395 = 11 + (n – 1) 12
->395 – 11 = 12 (n – 1)
32 = n – 1
n = 33
Sum =
=
= 6699
3, a, b, c are in A.P.
a – 3 = b – a (common diff.)
2a = b + 3
and 3, a – 1, b + 1 are in G.P.
a2 + 1 – 2a = 3b + 3
a2 – 8a + 7 = 0 &n
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