Question

Q3. In an A.P., the first term is 2 and the sum of the first five terms id one-fourth of the next five terms. Show that 20ᵗʰ term is -112.​

Answer 1

Step-by-step explanation:

Given :-

In an A.P., the first term is 2 and the sum of the first five terms is one-fourth of the next five terms.

To find :-

Show that 20ᵗʰ term is -112.

Solution:-

Given that

First term of an AP = 2

t1 = 2

We know that

The general form of an AP = t1,t1+d, t1+2d,...

We know that

nth term of an AP = tn = t1+(n-1)d

Now

The sum of first five terms

= t1 + t2 + t3 + t4 + t5

= t1 +( t1+d )+ (t1+ 2d) + (t1+3d) +( t1+4d)

=> 5 t1 + 10d

=> 5(2)+10d

=> 10+10d

The sum of first five terms = 10+10d -----(1)

and

The sum of next five terms

=> t6 +t7 + t8 + t9 +t10

=>( t1+5d) +(t1+6d) +(t1+7d) +(t1+8d)+(t1+9d)

=> 5t1 + 35d

=> 5(2) + 35d

=> 10+35d

The sum of next five terms = 10+35d ------(2)

Given that

The sum of the first five terms = one-fourth of the next five terms

=> 10+10d = (1/4)[10+35d]

=> 4(10+10d ) = (10+35d)

=> 40 + 40 d = 10 + 35 d

=> 40d -35d = 10-40

=> 5 d = -30

=> d = -30/5

=> d = -6

Common difference = -6

Now 20 th term of the AP

=> t 20

=> t1+(20-1)d

=> t1 +19d

=> 2+19(-6)

=> 2+(-114)

=> 2-114

=> -112

t20 = -112

Answer:-

The first term is 2 and the sum of the first five terms id one-fourth of the next five terms then 20ᵗʰ term is -112.

Used formulae:-

• The general form of an AP = t1,t1+d, t1+2d,...

• nth term of an AP = tn = t1+(n-1)d

Answer 2

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

Let first term is represented by a and common difference is represented by d of an AP series.

We have given that,

• First term of an AP, a = 2

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

↝ 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 first n terms.

• a is the first term of the sequence.

• n is the no. of terms.

• d is the common difference.

Tʜᴜs,

According to statement,

Sum of the first 5 terms is one fourth of sum of next 5 terms.

$\rm :\longmapsto\:S_5 = \dfrac{1}{4}(S_{10} - S_5)$

$\rm :\longmapsto\:4S_5 = S_{10} - S_5$

$\rm :\longmapsto\:5S_5 = S_{10}$

$\rm :\longmapsto\:5 \times \dfrac{5}{2} \bigg(2a + (5 - 1)d \bigg) = \dfrac{10}{2} \bigg(2a + (10 - 1)d \bigg)$

$\rm :\longmapsto\:5(2a + 4d) = 2(2a + 9d)$

$\rm :\longmapsto\:10a + 20d = 4a + 18d$

$\rm :\longmapsto\:2d = - 6a$

$\rm :\longmapsto\:d = - 3a$

$\rm :\longmapsto\:d = - 3 \times 2$

$\rm :\longmapsto\:d = - 6$

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,

$\rm :\longmapsto\:a_{20} = a + (20 - 1)d$

$\rm \: = \: \: a + 19d$

$\rm \: = \: \: 2 + 19 \times ( - 6)$

$\rm \: = \: \: 2 - 114$

$\rm \: = \: \: - 112$

$\bf\implies \:a_{20} \: = \: - \: 112$

${{\boxed{\bf{Hence, Proved}}}}$