Question

In an A.P., the sixth term is 27 and the eleventh term is 52. Find the 20th term.

Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

In an A.P., the 6th term is 27 and the 11th term is 52.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The 20th term.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

We know that;

• a is the first term.
• d is the common difference.
• n is term of A.P.

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

So;

$\dashrightarrow\sf{a_{6}=a+(6-1)d}\\\\\dashrightarrow\sf{a_{6}=a+5d}\\\\\dashrightarrow\sf{27=a+5d.........................(1)}$

&

$\dashrightarrow\sf{a_{11}=a+(11-1)d}\\\\\dashrightarrow\sf{a_{11}=a+10d}\\\\\dashrightarrow\sf{52=a+10d.......................(2)}$

Subtracting equation (1) from (2),we get;

$\mapsto\sf{\cancel{a}+10d\cancel{-a} -5d=52-27}\\\\\mapsto\sf{10d-5d=25}\\\\\mapsto\sf{5d=25}\\\\\mapsto\sf{d=\cancel{\dfrac{25}{5}}} \\\\\\\mapsto\sf{\green{d=5}}$

Putting the value of d in equation (1),we get;

$\mapsto\sf{27=a+5(5)}\\\\\mapsto\sf{27=a+25}\\\\\mapsto\sf{a=27-25}\\\\\mapsto\sf{\green{a=2}}$

The 20th term of an Arithmetic progression :

$\mapsto\sf{a_{20}=2+(20-1)5}\\\\\mapsto\sf{a_{20}=2+19\times 5}\\\\\mapsto\sf{a_{20}=2+95}\\\\\mapsto\sf{\green{a_{20}=97}}$

Thus;

The 20th term is 97 .

Answer 2

Given :-

In an A.P., the sixth term is 27 and the eleventh term is 52.

• a₆ = 27

• a₁₁ = 52

To find :-

The 20th term (a₂₀).

Solution :-

$\sf{a_6=a+5d=27}\ \:\:\:\:\:\:\:\:\:\:\:\:\:....(i)\\\sf{a_1_1=a+10d=52}\ \:\:\: \:\:\:\:\:....(ii)$

Subtracting eq.(i) from eq.(ii) :-

$\sf{a_1_1=a+10d=52}\ \:\:\:....(ii)\\-\\\sf{a_6=a+5d=27}\ \:\:\: \:\:\:\:\:....(i)\\---------\\5d=25$

So, we get d = 25/5 = 5

Putting the value of d = 5 in eq.(i) :

a + 5d = 27

⇒a + 5(5) = 27

⇒a = 27 - 25

⇒a = 2

We know,

a₂₀ = a + 19 d

⇒a₂₀ = 2 + (19*5)

⇒a₂₀ = 2 + 95

⇒a₂₀ = 97

∴ So, the 20th term of the A.P. is 97.