Question

If m times the mth term of an A.P. is eqaul to n times nth term then show that the (m + n)th term of the A.P. is zero. Solve the word problem

Answer 1

Step-by-step explanation:

If m times the mth term of an A.P. is equal to n times nth term

If first term of AP : a

common difference: d

than nth term of AP is

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

ATC

$m[a + (m - 1)d]= n[a + (n - 1)d] \\ \\ am + {m}^{2} d - dm = an + {n}^{2} d - dn \\ \\ am + {m}^{2} d - dm - an - {n}^{2} d + dn = 0 \\ \\ a(m - n) + ( {m}^{2} - {n}^{2} )d - d(m - n) = 0$

As we know that

${a}^{2} - {b}^{2} = (a + b)(a - b) \\ \\ a(m - n) + (m + n )(m - n)d - d(m - n) = 0$

By inspecting it is clear that (m-n) is common in all the terms

$(m - n)[a + (m + n)d - d] = 0 \\ \\ a + (m + n - 1)d = \frac{0}{m - n} \\ \\ a + (m + n - 1)d = 0$

The expression above is the expression of (m+n)th term of that AP

Hence

$T_{(m + n)} = 0$

Hope it helps you.

Answer 2

Answer:

Let the first term of AP = a

common difference = d

We have to show that (m+n)th term is zero or a + (m+n-1)d = 0

mth term = a + (m-1)d

nth term = a + (n-1) d

Given that m{a +(m-1)d} = n{a + (n -1)d}

⇒ am + m²d -md = an + n²d - nd

⇒ am - an + m²d - n²d -md + nd = 0

⇒ a(m-n) + (m²-n²)d - (m-n)d = 0

⇒ a(m-n) + {(m-n)(m+n)}d -(m-n)d = 0

⇒ a(m-n) + {(m-n)(m+n) - (m-n)} d = 0

⇒ a(m-n)  + (m-n)(m+n -1) d  = 0

⇒ (m-n){a + (m+n-1)d} = 0 

⇒ a + (m+n -1)d = 0/(m-n)

⇒ a + (m+n -1)d = 0

Proved!