1.

The sum of first 15th term of an arithmetic sequence is 570 and its 12th trem is 62What is the sum of 30th term

Answer»

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

We know that

↝ 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.

Now, Given that

Sum of first 15 terms of an AP = 570

\red{\rm :\longmapsto\:S_{15}\:=\dfrac{15}{2} \bigg(2 \:a\:+\:(15\:-\:1)\:d \bigg)}

\rm :\longmapsto\:570 = \dfrac{15}{2}(2a + 14d)

\rm :\longmapsto\:38 = \dfrac{2}{2}(a + 7d)

\bf\implies \:a + 7d = 38 -  -  - (1)

Also, we know that

↝ 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.

Now, Given that

\rm :\longmapsto\:a_{12}\:=62

\rm :\longmapsto\:a + (12 - 1)d = 62

\rm :\longmapsto\:a + 11d = 62 -  -  - (2)

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

\rm :\longmapsto\:4d = 24

\bf\implies \:d = 6

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

\rm :\longmapsto\:a + 7 \times 6 = 38

\rm :\longmapsto\:a + 42= 38

\rm :\longmapsto\:a = 38 - 42

\bf\implies \:a =  - 4

Now, Sum of first 30 terms is given by

\rm :\longmapsto\:S_{30}\:=\dfrac{30}{2} \bigg(2 \:( - 4)\:+\:(30\:-\:1)\:6 \bigg)

\rm :\longmapsto\:S_{30}\:=15 \bigg( -  \:8\:+\:(29)\:6 \bigg)

\rm :\longmapsto\:S_{30}\:=15 \bigg( -  \:8\:+\:174 \bigg)

\rm :\longmapsto\:S_{30}\:=15  \times 166

\bf :\longmapsto\:S_{30}\:=2490



Discussion

No Comment Found

Related InterviewSolutions