Question

The sum of the $4^{th}$ and $8^{th}$ term of an AP is 24 and the sum of the 6th and 10th term is 44. Find the first three terms of AP.

Answer 1

Hint: Here we write an equation for each of the given terms using the general formula of term of an AP. Add the corresponding equations of given terms and form two equations. Use a substituting method to find the value of first term and common difference of the AP. Calculate the first three terms of AP using the general formula of term of an AP.
* An arithmetic progression is a sequence of terms having common differences between them. If ‘a’ is the first term of an AP, ‘d’ is the common difference, then the $n^{th}$ term of an AP can be found as \[{a_n} = a + (n - 1)d\] .

Complete step-by-step answer:
Let us assume the first term of the AP as ‘a’ and the common difference as ‘d’.
Then nth term can be given by \[{a_n} = a + (n - 1)d\]...................… (1)
Put \[n = 4\]in equation (1)
\[ \Rightarrow {a_4} = a + (4 - 1)d\]
\[ \Rightarrow {a_4} = a + 3d\]...................… (2)
Put\[n = 8\]in equation (1)
\[ \Rightarrow {a_8} = a + (8 - 1)d\]
\[ \Rightarrow {a_8} = a + 7d\].......................… (3)
We are given that the sum of the 4th and 8th term of an AP is 24
\[ \Rightarrow {a_4} + {a_8} = 24\]
Substitute values from equations (2) and (3)
\[ \Rightarrow a + 3d + a + 7d = 24\]
Add like terms
\[ \Rightarrow 2a + 10d = 24\]......................… (4)
Put \[n = 6\]in equation (1)
\[ \Rightarrow {a_6} = a + (6 - 1)d\]
\[ \Rightarrow {a_6} = a + 5d\].....................… (5)
Put\[n = 10\] in equation (1)
\[ \Rightarrow {a_{10}} = a + (10 - 1)d\]
\[ \Rightarrow {a_{10}} = a + 9d\].................… (6)
We are given that the sum of the $6^{th}$ and $10^{th}$ term of an AP is 44
\[ \Rightarrow {a_6} + {a_{10}} = 44\]
Substitute values from equations (5) and (6)
\[ \Rightarrow a + 5d + a + 9d = 24\]
Add like terms
\[ \Rightarrow 2a + 14d = 44\]..................… (7)
Now we have to solve the equations (4) and (7)
From equation (4), \[2a + 10d = 24\]
On shifting value 10d to RHS
\[ \Rightarrow 2a = 24 - 10d\]
Substitute this value of 2a in equation (7)
\[ \Rightarrow 24 - 10d + 14d = 44\]
Shift all constant values to RHS of the equation
\[ \Rightarrow - 10d + 14d = 44 - 24\]
\[ \Rightarrow 4d = 20\]
Divide both sides by 4
\[ \Rightarrow \dfrac{{4d}}{4} = \dfrac{{20}}{4}\]
Cancel same factors from numerator and denominator
\[ \Rightarrow d = 5\]
Substitute the value of d in equation \[2a = 24 - 10d\]
\[ \Rightarrow 2a = 24 - 10 \times 5\]
\[ \Rightarrow 2a = 24 - 50\]
\[ \Rightarrow 2a = - 26\]
Divide both sides by 2
\[ \Rightarrow \dfrac{{2a}}{2} = \dfrac{{ - 26}}{2}\]
Cancel same factors from numerator and denominator
\[ \Rightarrow a = - 13\]
\[\therefore a = - 13,d = 5\]
Since a is the first term of AP, the value of first term is -13.
General formula for term of AP with \[a = - 13,d = 5\]is\[{a_n} = - 13 + (n - 1)5\]
Put \[n = 2\] in \[{a_n} = - 13 + (n - 1)5\]
\[ \Rightarrow {a_2} = - 13 + (2 - 1)5\]
\[ \Rightarrow {a_2} = - 13 + 5\]
\[ \Rightarrow {a_2} = - 8\]
Put \[n = 3\]in \[{a_n} = - 13 + (n - 1)5\]
\[ \Rightarrow {a_3} = - 13 + (3 - 1)5\]
\[ \Rightarrow {a_3} = - 13 + 10\]
\[ \Rightarrow {a_2} = - 3\]

\[\therefore \]First three terms of AP are -13, -8 and -3

Note: Alternate method:
Students can also find the three terms in the end by only using the common difference.
Since, ‘d’ is the common difference between the terms of AP.
We have \[a = - 13,d = 5\]
\[ \Rightarrow {a_2} = a + d\]
Substitute value of ‘a’ and ‘d’
\[ \Rightarrow {a_2} = - 13 + 5\]
\[ \Rightarrow {a_2} = - 8\]
Similarly,
\[ \Rightarrow {a_3} = {a_2} + d\]
Substitute value of ‘\[{a_2}\]’ and ‘d’
\[ \Rightarrow {a_3} = - 8 + 5\]
\[ \Rightarrow {a_3} = - 3\]
\[\therefore \]First three terms of AP are -13, -8 and -3