If (m +1)th term of an AP is twice the (n +1)th term, then the (3m +1)th term is twice the

(m + 1)th term of an AP = 2 * (n + 1)th term.

the (3m +1)th term is twice the​

We know that,

nth term of an AP – a(n) = a + (n - 1)d

Hence,

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

→ a + md = 2 (a + nd)

→ a + md = 2a + 2nd

→ md - 2nd = 2a - a

→ a = md - 2nd -- equation (1).

Now,

a(3m + 1) = 2 * a(n)

[ let the term be n]

→ a + (3m + 1 - 1)d = 2 * [ a + (n - 1)d ]

→ a + 3md = 2 * [ a + (n - 1)d ]

Substitute a value from equation (1).

→ md - 2nd + 3md = 2 * [ a + (n - 1)d ]

→ 4md - 2nd = 2 * [ a + (n - 1)d ]

→ 2(2md - nd) = 2 * (a + nd - d)

→ 2md - nd - a + d = nd

→ 2md - nd - (md - 2nd) + d = nd

[ From equation (1) ]

→ 2md - nd - md + 2nd + d = nd

→ md + nd + d = nd

→ (m + n + 1)d = n * d

→ (m + n + 1) = n

Hence, the (3m + 1)th term is twice the (m + n + 1)th term of given AP.

I hope it will help you.

Regards.

→ md - 2nd + 3md = 2 * [ a + (n - 1)d ]

→ 4md - 2nd = 2 * [ a + (n - 1)d ]

→ 2(2md - nd) = 2 * (a + nd - d)

→ 2md - nd - a + d = nd

→ 2md - nd - (md - 2nd) + d = nd

[ From equation (1) ]

→ 2md - nd - md + 2nd + d = nd

→ md + nd + d = nd

→ (m + n + 1)d = n * d

→ (m + n + 1) = n

Step-by-step explanation:

the probability is 0

1,589.63 in.^3

step-by-step explanation:

i hope this you : d