Question

$\huge \boxed { \tt \blue{\maltese \: Question}}$
In an A.P, 3rd term is 18 and 17th term is 60 then the 20th term is.. . . . .

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Answer 1

Concept Used :-

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ nᵗʰ term of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{a_n\:=\:a\:+\:(n\:-\:1)\:d}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

aₙ is the nᵗʰ term.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

Tʜᴜs,

$\large\underline{\sf{Solution-}}$

Let assume that

• First term of an AP series = a

• Common difference of an AP series = d

Given that, In an AP series,

$\rm :\longmapsto\:a_3 = 18$

$\bf\implies \:\boxed{ \tt{ \: a + 2d = 18}} - - - (1)$

and

$\rm :\longmapsto\:a_{17} = 60$

$\bf\implies \:\boxed{ \tt{ \: a + 16d = 60}} - - - (2)$

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

$\red{\rm :\longmapsto\:a + 16d - a - 2d = 60 - 18}$

$\red{\rm :\longmapsto\:14d = 42}$

$\red{\bf\implies \:\boxed{ \tt{ \: d \: = \: 3 \: \: }}} - - - - (3)$

On substituting the value of d, in equation (1), we get

$\purple{\rm :\longmapsto\:a + 2 \times 3 = 18}$

$\purple{\rm :\longmapsto\:a + 6 = 18}$

$\purple{\rm :\longmapsto\:a = 18 - 6}$

$\purple{\bf\implies \:\boxed{ \tt{ \: a \: = \: 12\: \: }}} - - - - (4)$

Now,

$\green{\rm :\longmapsto\:a_{20}}$

$\green{\rm \: = \: \:a + 19d}$

On substituting the values of a and d, we get

$\green{\rm \: = \: \:12+ 19 \times 3}$

$\green{\rm \: = \: \:12+ 57}$

$\green{\rm \: = \: \:69}$

$\green{\bf\implies \:\boxed{ \tt{ \: a_{20} = 69 \: }}}$

More to know :-

↝ Sum of n  terms of an arithmetic sequence is,

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{S_n\:=\dfrac{n}{2} \bigg(2 \:a\:+\:(n\:-\:1)\:d \bigg)}}}}}} \\ \end{gathered}$

Wʜᴇʀᴇ,

Sₙ is the sum of n terms of AP.

a is the first term of the sequence.

n is the no. of terms.

d is the common difference.

Answer 2

$\rm\underline{ Given:-}$

In the A.P 3rd term is 18

17th term is 60

$\rm \underline{To \: find:-}$

20th term of A.P

$\rm\underline{Solution:-}$

In an A.P

$a_n= a + (n - 1)d$

Where ,

• an = nth term
• a = first term
• n = no.of terms
• d = common difference

So,

3rd term is 18 i.e a3 = 18

$a_3 = a + (3 - 1)d$

$18 = a + 2d \: - - - 1$

Similarily,

17th term or A.P is 60 i.e a17 = 60

$a_{17} = a + (17 - 1)d$

$60= a + 16d - - 2$

Solving the two equations,

Subtracting the two equations

${a + 2d - a -16d = 18-60}$

${-14d = -42}$

${14d = 42}$

${d = 42/14}$

${d = 3}$

So, the common difference is 3 .

Substitute value of d in any equation

${18 = a+2d }$

${18 = a +2(3)}$

${18 = a+6}$

${18-6 = a}$

${a = 12}$

So, the first term is 12.

We need to find the 20th term of A.P

$a_n= a + (n - 1)d$

$a_{20} = a + (20 - 1)d$

$a_{20} = a + (19)d$

$a_{20} = 12 + (19)3$

$a_{20} = 12 + 57$

$a_{20} = 69$

So, the 20th term of A.P is 69

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