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

2 thoughts on “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”

1. [tex]\large\underline{\sf{Solution-}}[/tex]

   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,

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

   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.

   [tex]\rm :\longmapsto\:S_5 = \dfrac{1}{4}(S_{10} – S_5)[/tex]

   [tex]\rm :\longmapsto\:4S_5 = S_{10} – S_5[/tex]

   [tex]\rm :\longmapsto\:5S_5 = S_{10}[/tex]

   [tex]\rm :\longmapsto\:5 \times \dfrac{5}{2} \bigg(2a + (5 – 1)d \bigg) = \dfrac{10}{2} \bigg(2a + (10 – 1)d \bigg) [/tex]

   [tex]\rm :\longmapsto\:5(2a + 4d) = 2(2a + 9d)[/tex]

   [tex]\rm :\longmapsto\:10a + 20d = 4a + 18d[/tex]

   [tex]\rm :\longmapsto\:2d = – 6a[/tex]

   [tex]\rm :\longmapsto\:d = – 3a[/tex]

   [tex]\rm :\longmapsto\:d = – 3 \times 2[/tex]

   [tex]\rm :\longmapsto\:d = – 6[/tex]

   Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

   ↝ nᵗʰ term of an arithmetic sequence is,

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

   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,

   [tex]\rm :\longmapsto\:a_{20} = a + (20 – 1)d[/tex]

   [tex] \rm \: = \: \: a + 19d[/tex]

   [tex] \rm \: = \: \: 2 + 19 \times ( – 6)[/tex]

   [tex] \rm \: = \: \: 2 – 114[/tex]

   [tex] \rm \: = \: \: – 112[/tex]

   [tex]\bf\implies \:a_{20} \: = \: – \: 112[/tex]

   [tex]{{\boxed{\bf{Hence, Proved}}}}[/tex]

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2. 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

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