Question

If 'n' times the 'n' th term an Ap is equal to m times its 'm' th term, then(m+n) th therm is?​

Answer 1

$\red\bigstar$ EXPLANATION $\red\bigstar$

• Given

'n' times the 'n' th term = 'm' times 'm' th term

• To find

(m + n)th term of the AP

• Procedure

Let the first term of the AP be a

We know that,

$T_n\:of\:AP = a+(n-1)d$

Therefore,

The nth term of the AP = $a + (n-1)d$

The mth term of the AP = $a + (m-1)d$

The (m + n)th term of the AP = $a + (m+n-1)d$

n times nth term =   $n[a + (n-1)d] = an + n^2d - nd$

m times mth term = $m[a + (m-1)d] = am + m^2d - md$

Therefore,

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

We know that (m + n)th term = a + d(m + n -1)

Therefore,

(m + n)th term = 0

• Extra Information

i) $T_n\:of\:an\:AP=a+(n-1)d$

ii) $S_n\:of\:an\:AP=\frac{n[2a+(n-1)d]}{2} = \frac{n(a+l)}{2}$

iii) $T_n\:of\:GP = a{r}^{(n-1)$

iv) $S_n\:of\:GP=\frac{a({r}^{n} - 1)}{(r-1)} = \frac{a(1-{r}^n)}{(1-r)}$