Example 4

Transcript

Example 4, In an A.P. if mth term is n and the nth term is m, where m ≠ n, find the pth term. We know that an = a + (n – 1) d i.e. nth term = a + (n – 1) d Thus, mth term = am = a + (m – 1) d It is given that mth term is n a + (m – 1) d = n Also, it is given that nth term is m a + (n – 1) d = m First we find common difference, Subtracting (2) from (1) [a + (m – 1) d] – [a + (n – 1) d] = n – m a + (m – 1)d – a – (n – 1)d = n – m a – a + (m – 1)d –(n – 1)d = n – m (m – 1)d – (n – 1)d = n – m md – d – nd + d = n – m md – nd = n – m d(m – n) = n – m d = (𝑛 − 𝑚 )/(𝑚 −𝑛) d = ((𝑚 − 𝑛) )/(𝑚 −𝑛) × – 1 d = – 1 Now we have to calculate a Putting d = – 1 in (2) a + (n – 1) d = m a + (n – 1) × (-1) = m a – n + 1 = m a = m + n – 1 For pth term, we use the formula, an = a + (n – 1)d putting n = p, d = -1 and a = m + n – 1 ap = (m + n – 1) + ( p – 1) (–1) = m + n – 1 + (– p + 1) = m + n – 1 – p + 1 = m + n – p Thus, pth term = m + n – p