Question

can anuone solve ques 28, 29 and 30???

Answer 1

28) let a is the first term and d is the common difference of AP

a/c to question,

mth term = 1/n
a + (m-1)d = 1/n --------(1)

nth term = 1/m

a + (n - 1)d = 1/m ------------(2)

subtract equation (1) -(2)

(m- 1)d - (n- 1)d = 1/n - 1/m

d {m - n} = (m - n)/mn

d = 1/mn

put d = 1/mn in equation (1)

a = 1/n - (m - 1)1/mn = 1/mn

now ,

mnth term = a + (mn - 1)d
= 1/mn + (mn - 1)1/mn
= 1/mn + 1 - 1/mn = 1

so, mnth term = 1

29) pth term = q

a + (p - 1)d = q ---------(1)

qth term = p

a + (q - 1)d = p ---------(2)

subtract equation (1) - (2)

a + (p - 1)d - a -(q - 1)d = q - p = -(p - q)

(p - 1 - q + 1)d = -(p - q)

(p - q)d = - (p - q)


d = - 1

put d = - 1 in equation (1)

a = p + q - 1

now ,

nth term = a + (n - 1)d
= p + q -1 + (n -1) (-1)
=p + q -n

30)

according to question ,

p x pth term = q x qth term

p { a + ( p - 1)d } = q{ a + (q - 1)d}

a (p - q) + d { p^2 - p -q^2 + q } = 0

a (p - q) + d{ (p - q)( p + q) - (p- q)} =0

( p - q) {a + (p + q - 1)d } =0

a + (p + q - 1)d = 0 ---------(1)

now,

(p + q )th term = a + (p + q - 1)d

from equation (1)

(p + q )th term = a + (p +q - 1)d = 0

Answer 2

29. am = 1/n => a +(m-1)d

an= I/m => a+(n-1)d

amn = a+(mn-1)d=1 

subtract an from am

we get (m-1)d - (n-1)d = m-n/mn

by opening the brackets we get d = 1/mn

on substituting d in am we get a = 1/mn

now substitute in amn and we get the desired answer.

28. write the pth term and the qth terms then subtract both. you should get

(p-1)d-(q-1)=q-p

then d= -1

then find a which is q+p-1

sub this in the nth term you will get the answer.

30. write the given data as it is open the brackets group the values together

then you will get the p+qth term which equal to zero hence proved.