Question

the 4th term of an Ap is zero prove that the 25th term of the Ap is three time it's 11th term

Answer 1

Answer:

Given a^= 0 i.e

a+ 3d=0 ,=a= -3d ....1 n term of ap is given by an = a+(n - 1 ) d

a^ 11= a+ 10d = 7d fromm eq ..1

a^25 =a+24d= 21d=3×7d ...from 1

a^25 =3×a^11

Answer 2

Given,

4th term of AP = 0

To Find,

25th term of the AP = 3(11th term of the AP)

Solution,

From the formula for the nth term in AP,

$a_n = a + (n-1)d$

For the 4th term,

$a_4 = a + (4-1)d = a + 3d$

$a_4 =0 = a + 3d$ [equation 1]

For the 25th term of AP

$a_{25}= a + (25-1)d = a + 24d\\a_{25}= a + 24d = a + 3d + 21d \\a_{25}= 0 + 21d$[From equation 1, a +3d  = 0]

$a_{25} = 21d$                                [Equation2]

Similarly for  the 11th term in AP,

$a_{11} = a + 10d = a + 3d + 7d\\a_{11} = 0 + 7d \\a_{11} = 7d$   [Equation 3]

Multiply Equation 3 by 3, we get

$3 * a_{11} = 3 * 7d = 21d\\3* a_{11} = a_{21}$              [From equation 2]

The 25th term of the AP is three times it's 11th term. Hence proved.