Question

(A) Given $a = 3,n = 8,S = 192$. Find d
(B) Given $l = 28,S = 144$ and there are a total 9 terms. Find a

Answer 1

We solve for the values of ‘d’ and ‘a’ in the above given questions using the formula of sum of n terms of an Arithmetic progression. Substitute the given values in respective formulas and find the value of the missing term.

• An arithmetic progression is a sequence of terms having common differences between them. If ‘a’ is the first term of AP, ‘d’ is the common difference, then the ${n^{th}}$term of an AP can be found using the formula ${a_n} = a + (n - 1)d$.
• Sum of n terms of an AP is given by the formula${S_n} = \dfrac{n}{2}(a + l)$, where ‘a’ is the first term and ‘l’ is the last term of an AP. Sum can also be found using the formula ${S_n} = \dfrac{n}{2}(a + (n - 1)d)$

Complete step-by-step solution:
(A) We are given an Arithmetic Progression.
First term of AP,$a = 3$
Number of terms of AP,$n = 8$
Sum of n terms of AP,$S = 192$
We have to find the value of the common difference ‘d’ of the AP.
Since we are given the values of ‘a’, ‘n’ and ‘S’; we will use the formula of sum of n terms of AP i.e. ${S_n} = \dfrac{n}{2}(a + (n - 1)d)$
Substitute the values of $a = 3,n = 8,S = 192$in the formula of sum of n terms
$\Rightarrow 192 = \dfrac{8}{2}(3 + (8 - 1)d)$
Cancel same factors from numerator and denominator in RHS of the equation
$\Rightarrow 192 = 4(3 + 7d)$
Divide both sides of the equation by 4
$\Rightarrow \dfrac{{192}}{4} = \dfrac{{4(3 + 7d)}}{4}$
Cancel the same factors from numerator and denominator on both sides of the equation.
$\Rightarrow 48 = 3 + 7d$
Shift all constant values to one side of the equation
$\Rightarrow 7d = 48 - 3$
$\Rightarrow 7d = 45$
Divide both sides by 7
$\Rightarrow \dfrac{{7d}}{7} = \dfrac{{45}}{7}$
Cancel the same factors from the numerator and denominator on both sides of the equation.
$\Rightarrow d = \dfrac{{45}}{7}$
$\Rightarrow d = 6.42$
$\therefore$Value of ‘d’ is 6.42.
(B) We are given an Arithmetic Progression.
Last term of AP, $l = 28$
Number of terms of AP, $n = 9$
Sum of n terms of AP, $S = 144$
We have to find the value of the first term ‘a’ of the AP.
Since we are given the values of ‘l’, ‘n’ and ‘S’; we will use the formula of sum of n terms of AP i.e. ${S_n} = \dfrac{n}{2}(a + l)$
Substitute the values of $l = 28,n = 9,S = 144$ in the formula of sum of n terms
$\Rightarrow 144 = \dfrac{9}{2}(a + 28)$
Cross multiply 2 from denominator of RHS to numerator of LHS
$\Rightarrow 144 \times 2 = 9(a + 28)$
$\Rightarrow 288 = 9(a + 28)$
Divide both sides of the equation by 9
$\Rightarrow \dfrac{{288}}{9} = \dfrac{{9(a + 28)}}{9}$
Cancel the same factors from numerator and denominator on both sides of the equation.
$\Rightarrow 32 = a + 28$
Shift all constant values to one side of the equation
$\Rightarrow a = 32 - 28$
$\Rightarrow a = 4$
$\therefore$Value of ‘a’ is 4.

Note: Students many times make mistakes in calculations where they forget to change the sign of the value when shifting it to the other side of the equation. The sign changes from positive to negative and vice versa when shifting values from one side to other side. Also, many students are unfamiliar with what ‘l’ indicates. ‘l’ indicates the last term of AP, you can easily remember it as the initial of the word ‘Last’.