Question

if (m + 1)th term of an ap is twice the (n + 1)th term prove that (3m+ 1)th term is twice the (M + n + 1)th term​

Answer 1

Answer:

Proved!

Step-by-step explanation:

Given that :

If (m + 1)th term of an ap is twice the (n + 1)th term.

To Prove :

(3m+ 1)th term is twice the (M + n + 1)th term​.

Solution :

$\textbf{\underline{\underline{Consider as :-}}}$

The first term of AP = a.

The common difference of AP = d.

According to the question :

$\sf \implies (m+1) th\;term =2 (n+1) th\;term\\ \\\implies a+(m+1-1) d =2 {a+(n+1-1) d }\\ \\\implies a+md=2a+2nd\\ \\\implies (m-2n)d=a ......Eq(1)\\ \\ \text{Now}\\ \\\text{LHS}} =(3m+1) th\;term\\ \\ \implies a+(3m+1-1) d\\ \\\implies (m-2n) d+3md\\ \\\implies 2 (2m-n) d\\ \\\text RHS =2 (m+n+1) th\;term\\ \\\implies 2 {a+(m+n+1-1) d}\\ \\\implies 2 {a+(m+n) d}\\ \\\implies 2 {(m-2n)d +(m+n) d}\\ \\\implies 2 (2m-n) d$

Hence, LHS = RHS.

Means,  

(3m+ 1)th term is twice the (M + n + 1)th term​

Proved!

Answer 2

$\huge\red{Answer}$

Let's assume that a and d are 1st term and common difference respectively of the A.P

So , according to the question

(m+1)th term = 2( n+1)th term .

Using the formula .

an = a+(n-1)d

m+1 = 2 (n+1)

a + [(m+1)-1]d =. 2 {a+[(n+1)-1]d

a+(m+1-1)d. = 2[a+(n+1-1)]d

a+md. = 2a +2nd

a. = d ( m-2n) .........eq(1)

Now , For 3m+1 th term

= a + [(3m+1)-1]d

= a + 3md

Putting the value of a from 1st eq.

= d(m-2n) +3md

= md -2nd +3md

= 4md -2nd .......eq(2)

As 3m+1 is equal to twice of ( m+n+1)th term.

Now , For (m+n+1) th term

= 2{a+[(m+n+1)-1]d

= 2[a+(m+n)d]

Again putting the value of a from 1st equation.

= 2[(m-2n)d +(m+n)d]

= 2(md-2nd +md +nd )

= 2(2md -nd)

= 4md-2nd...........eq(3)

So from above it is clear that eq 2 nd and 3rd are equal.

4md-2nd = 4md-2nd

hence proved.