Question

Write an A.P., whose
(a) 10th term is 52 and 16th term is 82

Answer 1

Answer:

a+9d=52 eq.1

a+15d=82 eq.2

Then by using elimination method:

a+9d=52

a+15d=82

- - -

0-6d=-30

6d=30

d=5

Put the value of g in eq.1

a+9d=52

a+9 (5)=52

a=52-45

a=7

Therefore A.P is 7,12,17,22,27,32,37,42,47,52,57,62,67,72,77,82,85..........n

Step-by-step explanation:

Answer 2

Answer :

$\large{\dag\:\:\boxed{\bf{A.P.\:\: :-\:7\:,\:12\:,\:17\:,...}}\:\:\dag}$

Explanation :

Given :–

• a₁₀ = 52
• a₁₆ = 82

To Find :–

• Sequence of this A.P. (Arithmetic Progression)

Formula Applied :–

• $\boxed{\bf{\star\:\:a_n\:=\:a\:+\:(n\:-\:1)d\:\:\star}}$

Solution :–

We have ,

$\rightarrow\sf{a_{10}\:=\:52}$

$\rightarrow\sf{a_{10}\:=\:a\:+\:(10\:-\:1)d}$

$\rightarrow\sf{52\:=\:a\:+\:9d}\:\:-----\:\bf{(1)}$

We also have :

$\rightarrow\sf{a_{16}\:=\:82}$

$\rightarrow\sf{a_{16}\:=\:a\:+\:(16\:-\:1)d}$

$\rightarrow\sf{82\:=\:a\:+\:15d}\:\:-----\:\bf{(2)}$

Subtracting Equation(2) from Equation(1) :-

$\rightarrow\sf{82\:-\:52\:=\:(a\:+\:15d)\:-\:(a\:+\:9d)}$

$\rightarrow\sf{30\:=\:a\:+\:15d\:-\:a\:-\:9d}$

$\rightarrow\sf{30\:=\:6d}$

$\rightarrow\sf{d\:=\:\dfrac{30}{6} }$

$\rightarrow\boxed{\bf{d\:=\:5 }}$

Putting this Value of 'd' in Equation(1) :-

$\rightarrow\sf{52\:=\:a\:+\:9(5)}$

$\rightarrow\sf{52\:=\:a\:+\:45}$

$\rightarrow\sf{a\:=\:52\:-\:45}$

$\rightarrow\boxed{\bf{a\:=\:7}}$

Now we have to find terms of this A.P.  :-

★ First Term :-

⇒ a₁ = 7

★ Second Term :-

⇒ a₂ = a + (2 - 1)d

⇒ a₂ = 7 + 5

⇒ a₂ = 12

★ Third Term :-

⇒ a₃ = a + (3 - 1)d

⇒ a₃ = 7 + 2(5)

⇒ a₃ = 7 + 10

⇒ a₃ = 17

∴ The sequence of the A.P. will be 7 , 12 , 17 ,...