Question

The 10th term and the 18th term of an A.P. are 41 and 73 respectively. Find the following: a) 100th term b) Difference between 100th term and 50th term​

Answer 1

Answer:

Asked

➝ 100th term

➝ Difference between 100th term and 50th term

Solution

Given

10th term ➝ 41

18th term ➝ 73

We can also write it as

a + 9d = 41. ( 10th term ). (i)

a + 17d = 73 (18th term ) (ii)

Now subtracting

a + 9d = 41

- a + 17d = 73

(-) (-) (-)

________________________

- 8d = - 32

d = 32

8

d = 4

Now putting the value of d in (i)

a + 9×4 = 41

a = 41 - 36

a = 5

100th term

100th term ➝ a + 99d

Now putting the value of a and d

➝ 5 + (99 × 4)

➝ 5 + 396

➝ 401

Difference between 100th term and 50th term

╰➝ 100th term ➝ 401

╰➝ 50th term ➝ a + 49d

➝ 5 + 49 × 4

➝ 5 + 196

➝ 201

Difference

╰➝ 100th - 50th

➝ 401 - 201

➝ 200

How it helps you dear (◕ᴗ◕✿)

Answer 2

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

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,

According to statement, it is given that

$\rm :\longmapsto\:a_{10} = 41$

$\rm :\longmapsto\:a + (10 - 1)d = 41$

$\rm :\longmapsto\:a + 9d = 41 - - - (1)$

Also,

$\rm :\longmapsto\:a_{18} = 73$

$\rm :\longmapsto\:a + (18 - 1)d = 73$

$\rm :\longmapsto\:a + 17d = 73 - - - (2)$

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

$\rm :\longmapsto\:17d - 9d = 73 - 41$

$\rm :\longmapsto\:8d = 32$

$\rm :\implies\:\red{\boxed{ \bf{ \: \: \: d \: = \: 4 \: \: \: }}}$

On substituting the value of d = 4, in equation (1), we get

$\rm :\longmapsto\:a + 9 \times 4 = 41$

$\rm :\longmapsto\:a + 36 = 41$

$\rm :\longmapsto\:a = 41 - 36$

$\rm :\implies\:\red{\boxed{ \bf{ \: \: \: a \: = \: 5\: \: \: }}}$

Now,

$\rm :\longmapsto\:a_{100}$

$\rm \: = \: a + (100 - 1)d$

$\rm \: = \: a + 99d$

$\rm \: = \: 5 + 99 \times 4$

$\rm \: = \: 5 +396$

$\rm \: = \: 401$

$\rm :\implies\:\red{\boxed{ \bf{ \: \: \: a_{100}\: = \: 401\: \: \: }}}$

Now,

$\rm :\longmapsto\:a_{100} - a_{50}$

$\rm \: = \: \bigg(a + (100 - 1)d\bigg) - \bigg(a + (50 - 1)d\bigg)$

$\rm \: = \: \bigg(a + 99d\bigg) - \bigg(a + 49d\bigg)$

$\rm \: = \: 50d$

$\rm \: = \: 50 \times 4$

$\rm \: = \: 200$

$\rm :\implies\:\red{\boxed{ \bf{ \: \: \: a_{100}\: - \: a_{50} = \: 200\: \: \: }}}$