Question

Prove the statement given below

In an A.P
$\sf t_{m+n}+t_{m-n}=2t_m$

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Answer 1

Answer:

tn=a+(n-1)d

t(m+n)+t(m-n) =a+(m+n-1)d+a+(m-n-1)d

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

=2a+d(2m-2)

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

=2tm

Answer 2

Step-by-step explanation:

Given Question :-

Prove that ,In an AP t m+n + t m-n = 2 tm

Solution :-

Let the first term be t1 in an AP

The common difference = d

We know that

The general term of an AP is denoted by tn and it is defined by tn = t1 +(n-1) d

Here , n is the number of terms

Now,

nth term = tn = t1 +(n-1) d

(m+n)th term = t m+n

=>t1 + (m+n-1)d

=>t1+md+nd-d

t m+n = t1+md+nd-d --------(1)

(m-n)th term = t m-n

=>t1 +(m-n-1)d

=>t1+md-nd-d

t m-n = t1+md-nd-d --------(2)

Now

LHS :-

t m+n + t m-n

from (1)&(2)

=>(t1+md+nd-d)+(t1+md-nd-d )

=>t1 + md +nd -d + t1 +md -nd -d

=>(t1+t1)+(md+md)+(nd-nd)+(-d-d)

=>2t1+2md+0+(-2d)

=>2t1+2md-2d

=>2(t1+md-d)

=>2[t1+(m-1)d]

=>2(t m)

=>2(mth term of the AP)

=>RHS

t m+n + t m-n= 2 t m

Answer:-

t m+n + t m-n= 2 t m

Hence ,proved

Used formula:-

• t1 is the first term and d is the common difference of an AP then the general or nth term of an AP is denoted by tn and defined by tn = t1+(n-1)d .