Question

13. The 11th term and 31th term of an A.P. are 25 and 65 respectively. Find the nth term of the A.P.​

Answer 1

Step-by-step explanation:

$\bf{\huge\mathcal{\boxed{\rm{\mathfrak\blue{Hey Mate:}}}}}$

• T11 = 25 = a + (11-1)d
• Or, 25 = a + 10d .....(1)

• T31 = a + (31-1)d = 65
• Or, 65 = a + 30d ......(2)

(2)-(1) :

20d = 40

d = 40/20 = 2

• a = 65-30d = 65 - 30(2)
• = 5

• Tn = 5 + (n-1)(2)
• Tn = 5 + 2n -2
• Tn = 3+2n

Answer 2

Given :

The 11th term and 31th term of an A.P. are 25 and 65.

$a_{11}\:=\:25$

$a_{31}\:=\:65$

Find :

nth term of an AP

Solution :

→ $a_{11}\:=\:25$

→ $a\:+\:(11\:-\:1)d\:=\:25$

→ $a\:+\:10d\:=\:25$

→ $a\:=\:25\:-\:10d$ ____ (eq 1)

Similarly,

→ $a_{31}\:=\:65$

→ $a\:+\:(31\:-\:1)d\:=\:65$

→ $a\:+\:30d\:=\:65$

→ $a\:=\:65\:-\:30d$ ____ (eq 2)

From (eq 1) and (eq 2)

→ 25 - 10d = 65 - 30d

→ 25 - 65 = - 30d + 10d

→ - 40 = - 20d

→ d = 2

Put value of d in (eq 1)

→ a = 25 - 10(2)

→ a = 25 - 20

→ a = 5

We have to find the nth term of an AP

So,

$a_{n}\:=\:a\:+\:(n\:-\:1)d$

From above calculations we have a = 5 and d = 2

→ $a_{n}\:=\:5\:+\:(n\:-\:1)2$

→ $a_{n}\:=\:5\:+\:2n\:-\:2$

→ $a_{n}\:=\:3\:+\:2n$

∴ 3 + 2n is the nth term of an AP.