Question

If (m+1)th term of an A.P. is twice the (n+1)th term, prove that (3m+1)th term is twice (m+n+1)th term.

Answer 1

Answer:

It is proved that , (3 m + 1) th term of an A.P is twice the          (m + n+ 1) th term .

Step-by-step explanation:

Given as :

For Arithmetic progression

(m+1) th term of an A.P. is twice the (n+1) th term

Since , nth term of A.P = $t_n$ = a + (n - 1) d

where a is first term

And d is common difference

Now,

As , $t_n$ = a + (n - 1) d

So, $t_m_+_1$ = a + ( m + 1 - 1 )d

i.e  $t_m_+_1$ = a + m d                      

Again

As , $t_n$ = a + (n - 1) d

So, $t_n_+_1$ = a + ( n + 1 - 1 )d

i.e  $t_n_+_1$ = a + n d                      

According to question

(m+1) th term = 2 × (n+1) th term

i.e  $t_m_+_1$ = 2 × $t_n_+_1$

Or, a + m d = 2 × ( a + n d)

Or, a + m d = 2 a + 2 n d

Or, m d - 2 n d = 2 a - a

i.e  a = (m - 2 n ) d               ......1

To prove

(3 m + 1) th term of an A.P. is twice the (m + n+ 1) th term

So, As , $t_n$ = a + (n - 1) d

So, $t_3_m_+_1$ = a + ( 3 m + 1 - 1 )d

i.e $t_3_m_+_1$  = a + 3 m d        

Put the value a from eq 1  

$t_3_m_+_1$  = (m - 2 n )d + 3 m d    

or, $t_3_m_+_1$  = m d - 2 n d + 3 m d

∴  $t_3_m_+_1$  = 4 m d - 2 n d

i.e $t_3_m_+_1$  = 2 ( 2 m d - n d )              ........2

And

So, As , $t_n$ = a + (n - 1) d

So, $t_m_+_n_+_1$ = a + ( m + n + 1 - 1 ) d

i.e  $t_m_+_n_+_1$ = a + (m + n) d  

put the value of a from eq 1

$t_m_+_n_+_1$ =  (m - 2 n )d  + (m + n) d  

Or, $t_m_+_n_+_1$ = m d - 2 n d + m d + n d

Or, $t_m_+_n_+_1$ = 2 m d - n d                            ..........3

From eq 2 and eq 3

$t_3_m_+_1$  = 2  $t_m_+_n_+_1$

i.e  $t_3_m_+_1$  = twice of  $t_m_+_n_+_1$

Hence It is proved that , (3 m + 1) th term of an A.P is twice the (m + n+ 1) th term . Answer

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!