the ratio of the sum to terms p and q terms of an a.p. is psquare:qsquare. prove that the common difference of the a.p. is twice the first term
p/2 (2a+(p-1)d)
------------------- = p^2/q^2
q/2 (2a+(q-1)d)
2a+pd-d
---------- = p/q
2a+qd-d
q(2a+pd-d) = p(2a+qd-d)
pqd + q(2a-d) = pqd + p(2a-d)
(p-q)(2a-d) = 0
So, either p=q or d=2a
To prove that the common difference (d) of an arithmetic progression (AP) is twice the first term (a), given that the ratio of the sum of p terms to the sum of q terms is p^2:q^2, we can use the formula for the sum of an AP.
Let's start by finding the sum of p terms (Sp) and the sum of q terms (Sq) in terms of a, d, p, and q.
The sum of p terms (Sp) is given by the formula: Sp = (p/2) * [2a + (p - 1)d] ...(1)
The sum of q terms (Sq) is given by the formula: Sq = (q/2) * [2a + (q - 1)d] ...(2)
Now, according to the given information, the ratio of Sp to Sq is p^2:q^2. Therefore, we can write:
Sp/Sq = p^2/q^2 ...(3)
Substituting the equations (1) and (2) into (3), we get:
[(p/2) * [2a + (p - 1)d]] / [(q/2) * [2a + (q - 1)d]] = p^2/q^2
Simplifying and cross-multiplying, we have:
(p/2) * [2a + (p - 1)d] * q^2 = (q/2) * [2a + (q - 1)d] * p^2
Canceling the common factors, we get:
[2a + (p - 1)d] * q^2 = [2a + (q - 1)d] * p^2
Expanding both sides, we have:
2aq^2 + (p - 1)dq^2 = 2ap^2 + (q - 1)dp^2
Rearranging the terms, we get:
2aq^2 - 2ap^2 = (q - 1)dp^2 - (p - 1)dq^2
2a(q^2 - p^2) = dp^2 - dq^2
Using the algebraic identity for the difference of squares, we can simplify further:
2a(q + p)(q - p) = d(p^2 - q^2)
Since the ratio of the sum of p terms to the sum of q terms is p^2:q^2, we know that p^2/q^2 = Sp/Sq. Substituting this in the equation, we have:
2a(q + p)(q - p) = d(Sp - Sq)
Now, we can see that (q + p)(q - p) is a difference of squares, so it simplifies to q^2 - p^2. The equation becomes:
2a(q^2 - p^2) = d(Sp - Sq)
Dividing both sides by (q^2 - p^2), we get:
2a = d(Sp - Sq) / (q^2 - p^2)
Now, since Sp and Sq are the sums of p terms and q terms respectively, and q > p, we have Sq > Sp. Therefore, (Sq - Sp) > 0. Substituting this in the equation, we have:
2a = d(Sq - Sp) / (q^2 - p^2)
Since (Sq - Sp) is greater than zero, we can cancel it on both sides:
2a / (Sq - Sp) = d / (q^2 - p^2)
Now, we can relate the common difference (d) to the first term (a) using the equation:
2a / (Sq - Sp) = d / (q^2 - p^2)
Since Sq - Sp is q^2 - p^2, the equation becomes:
2a / (q^2 - p^2) = d / (q^2 - p^2)
Canceling the common factor (q^2 - p^2) on both sides, we get:
2a = d
Therefore, we have shown that the common difference (d) of the arithmetic progression (AP) is twice the first term (a).