Question

Find the ap if the 6th term of the ap is 19 and the 16th term is 15 more than the 11th term

Answer 1

We know that,

• 6th term = a + 5d
• 16th term = a + 15d
• 11th term = a + 10d

→ a + 5d = 19 [Given]

→ a + 15d = 15 + a + 10d [Given]

On solving we get that,

→ 5d = 15

→ d = 3

By substituting values we get,

→ a + 5d = 19

→ a + 5(3) = 19

→ a = 4

Hence Arithmetic Progression is:

→ 4, 7, 10, 13, 16, 19, 22, 25...

Answer 2

$\underline{\textbf{\large{Answer:}}}$

here, we have given,

▶6th term of the ap is 19

t6 = 19

▶16th term is 15 more than the 11th term

t16 = 15 + t11

we know the formula to find the terms in an AP,

tn = a + (n - 1) d

therefor,

t6 = a + (6 - 1) d

t6 = a + 5d

a + 5d = 19 ___________(1)

Now,

→ t16 = 15 + t11

→ t16 = 15 + ( a + (11 - 1)d)

→ (a + (16 - 1) d) = 15 + ( a + (11 - 1)d)

→ (a + 15 d) = 15 + a + 10d

→ 15d = 15 + 10d

→ 15d - 10d = 15

→ 5d = 15

$\boxed{\underline{\textbf{\large{ d = 3}}}}$

put the value of d in equation (1)

→ a + 5d = 19

→ a + 5(3) = 19

→ a + 15 = 19

→ a = 19 - 15

$\boxed{\underline{\textbf{\large {a = 4 }}}}$

the terms are,

t1 = a = 4

t2 = t1 + d = 4 + 3 = 7

t3 = t2 + d = 7 + 3 = 10

t4 = t3 + d = 10 + 3 = 13

t5 = t4 + d = 13 + 3 = 16

therefor the AP is 4,7,10,13,16.....