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If ‘a’ is the first term and ‘d’ is the common difference of the arithmetic progression, then its nth term is given by an = a+(n-1)d
The sum, Sn of the first ‘n’ terms of the A.P. is given by Sn = n/2 [2a + (n-1)d]
If Sn is the sum of n terms of an A.P. whose first term is ‘a’ and last term is ‘l’,Sn = (n/2)(a + l)
If common difference is d, number of terms n and the last term l, then Sn = (n/2)[2l-(n -1)d]
If a fixed number is added or subtracted from each term of an A.P., then the resulting sequence is also an A.P. and it has the same common difference as that of the original A.P.
If each term of A.P is multiplied by some constant or divided by a non-zero fixed constant, the resulting sequence is an A.P. again.
If a1, a2, a3, …, an andb1, b2, b3, …, bn, are in A.P. then a1+b1, a2+b2, a3+b3, ……, an+bn and a1–b1, a2–b2, a3–b3, ……, an–bn will also be in A.P.
Suppose a1, a2, a3, ……,an are in A.P. then an, an–1, ……, a3, a2, a1 will also be in A.P.
If nth term of a series is tn = An + B, then the series is in A.P.
If a1, a2, a3, ……, an are in A.P., then a1 + an = a2 + an–1 = a3 + an–2 = …… and so on.
In order to assume three terms in A.P. whose sum is given, they should be assumed as a-d, a, a+d.
Four terms of the A.P. whose sum is given should be assumed as a-3d, a-d, a+d, a+3d
Five convenient numbers in A.P. a–2b, a–b, a, a+b, a+2 b.
In general, we take a – rd, a – (r – 1)d, …., a – d, a, a + rd in case we have to take (2r + 1) terms in an A.P.
Likewise, any 2r terms of an A.P. should be assumed as: a – (2r-1)d, a – (2r – 3)d, …., a – d, a, a + d, ………….. , a+(2r-3)d, a + (2r-1)d.
The arithmetic mean of two numbers ‘a’ and ‘b’ is (a+b)/2.
The terms A1, A2, ….. , An are said to be arithmetic means between a and b if a, A1, A2, ….. , An, bis an A.P.
Clearly, ‘a’ is the first term, ‘b’ is the (n+2)th term and ‘d’ is the common difference. Then, we have b = a+(n+2-1)d = a+(n+1)d
Hence, this gives ‘d’ = (b-a)/(n+1)
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Solved Examples on Arithmetic Progression...