If the sum of an A.P IS THE SAME FOR P AS for q terms show that its sum for p+q terms is zero.
show the steps.
show the steps.
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i) Let the first term of the AP be 'a' and its common difference be 'd'
ii) Sum of p terms is Sp = (p/2){2a + (p - 1)d} ------ (1)
iii) Sum of q terms is Sq = (q/a){2a + (q - 1)d} ------ (2)
iv) Since given Sp = Sq, equating (1) & (2)
2ap + p²d - pd = 2aq + q²d - qd
Rearranging and grouping, 2a(p - q) + d(p² - q²) - d(p - q) = 0
==> (p - q)[2a + {(p + q) - 1}d] = 0
Since p & q are different, (p - q) is not equal to zero.
So from the above only, [2a + {(p + q) - 1}d] = 0 -------- (3)
v) Sum of (p + q) terms: S₍p+q₎ = {(p+q)/2}*[2a + {(p + q) - 1}d]
But from (3), [2a + {(p + q) - 1}d] = 0
Hence S₍p+q₎ = {(p+q)/2}*(0) = 0
Thus it is proved that sum to (p + q) terms is zero.
ii) Sum of p terms is Sp = (p/2){2a + (p - 1)d} ------ (1)
iii) Sum of q terms is Sq = (q/a){2a + (q - 1)d} ------ (2)
iv) Since given Sp = Sq, equating (1) & (2)
2ap + p²d - pd = 2aq + q²d - qd
Rearranging and grouping, 2a(p - q) + d(p² - q²) - d(p - q) = 0
==> (p - q)[2a + {(p + q) - 1}d] = 0
Since p & q are different, (p - q) is not equal to zero.
So from the above only, [2a + {(p + q) - 1}d] = 0 -------- (3)
v) Sum of (p + q) terms: S₍p+q₎ = {(p+q)/2}*[2a + {(p + q) - 1}d]
But from (3), [2a + {(p + q) - 1}d] = 0
Hence S₍p+q₎ = {(p+q)/2}*(0) = 0
Thus it is proved that sum to (p + q) terms is zero.