Question

Solution needed!!!! ​

Answer 1

Answer:

this question is answer.....

Answer 2

$\small{\sf{\underline{Question:−}}}$

If m times the mth term of an A.P. is equal to n times the nth term, show that (m+n)th term of the A.P. is zero.

$\large{\underline{\sf{Solution:-}}}$

Using the given relation,

• $ma_m = na_n$

As

• $a_b = a+(b-1)d$

So,

= $m[a+(m-1)d] = n[a+(n-1)d]$

Therefore,

= $am + m(m-1)d = an + n(n-1)d$

= $am + (m^2 - m)d = an + (n^2 - n)d$

= $am - an + (m^2 - m)d - (n^2 - n)d = 0$

= $a(m-n) + d(m^2 - m - n^2 + n) = 0$

As we know that,

• $a^2 - b^2 = (a+b)(a-b)$

= $a(m-n) + d[(m+n)(m-n) - (m-n)] = 0$

= $a(m-n) + d(m-n)(m+n - 1) = 0$

Dividing the whole equatiom by m - n :

= $\dfrac{a(m-n) + d(m-n)(m+n-1)}{m-n} = 0$

= $a + (m+n-1)d=0$

Using the same result for any term, we get:

• $a_{m+n} = 0$

Thus the required answer is proved!

Hope it helps❤