Question

the 12th term of an a.p is 9th times the second show that the 17th term is five times the fourth

Answer 1

A/C to question,
12th term in AP = 9 × 2nd term in AP
a + ( 12 - 1)d = 9[ a + (2 - 1)d ]
Here, a is the first term and d is the common difference in ap
a + 11d = 9a + 9d
8a = 2d ⇒d = 4a -----(1)

Now, we have to prove ,
7th term in AP = 5 × 4th term in AP
LHS = 17th term in AP
= a + (17 - 1)d
= a + 16d
= a + 16 × 4a = 65a [ from equation (1) ]

RHS = 5 × 4th term in AP
= 5 [a + (4 - 1)d ]
= 5 [ a + 3d]
= 5 [ a + 3 × 4a ] = 65a [ from equation (1) ]

Here, LHS = RHS
Hence it is proved that 17th term in AP = 5 × 4th term in AP

Answer 2

Answer:

A/C to question,

12th term in AP = 9 × 2nd term in AP

a + ( 12 - 1)d = 9[ a + (2 - 1)d ]

Here, a is the first term and d is the common difference in ap

a + 11d = 9a + 9d

8a = 2d ⇒d = 4a -----(1)

Now, we have to prove ,

7th term in AP = 5 × 4th term in AP

LHS = 17th term in AP

= a + (17 - 1)d

= a + 16d

= a + 16 × 4a = 65a [ from equation (1) ]

RHS = 5 × 4th term in AP

= 5 [a + (4 - 1)d ]

= 5 [ a + 3d]

= 5 [ a + 3 × 4a ] = 65a [ from equation (1) ]

Here, LHS = RHS

Hence it is proved that 17th term in AP = 5 × 4th term in AP