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

Hi dear❤️❤️

Given (m+1)th term = 2(n+1) th term . Let the first term of AP be a and common difference be d . According to question (m+1) the term = 2(n+1)the term

a+md = 2(a+nd).

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

now,we have to prove. (3m+1) th term. = 2(m+n+1) th term

LHS= (3m+1) th term. = a + (3m+1-1)d= a+3md = (m-2n )d +3md = 4md-2nd= 2d( 2m-n). [ For a use equation ....1]

RHS = 2(m+n+1) th term = 2[a+(m+n+1-1) d ] = 2[ a+ md+nd] = 2[ md-2nd + md+nd]. = 2[2md-nd]. = 2d[2m-n] = LHS. HENCE , PROVED ✌️✌️✌️❤️❤️

Answer 2

Step-by-step explanation:

Note : nth term of an AP is an = a + (n - 1) * d.

(i)

Given that (m + 1)th term of an AP is twice the (n + 1)th term.

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

⇒ a + md = 2[a + nd]

⇒ a + md = 2a + 2nd

⇒ -a = 2nd - md

⇒ a = md - 2nd

⇒ a = (m - 2n)d

(ii)

Consider, (3m + 1)th term:

⇒ a + (3m + 1 - 1) * d

⇒ a + 3md

⇒ (m - 2n) * d + 3md

⇒ md - 2nd + 3md

⇒ 4md - 2nd

⇒ (4m - 2n)d   ------ (1)

(iii)

Consider, (m + n + 1)th term.

⇒ a + (m + n + 1 - 1) * d

⇒ a + md + nd

⇒ (m - 2n) * d + md + nd

⇒ md - 2nd + md + nd

⇒ 2md - nd

⇒ (2m - n)d   ---- (2)

Therefore, From (1) & (2),

We can say that (3m + 1)Th term is twice the (m + n + 1)th term.

Hope it helps!