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Q3(viii):
In an AP: (viii) given $a_n = 4, d = 2, S_n = –14$, find $n$ and $a$.

Solution :

Given:

The $n^{th}$ term of the Arithmetic Progression (AP), $a_n = 4$.

The common difference of the AP, $d = 2$.

The sum of the first $n$ terms of the AP, $S_n = -14$.

To find:

The number of terms, $n$, and the first term, $a$.

Step 1: Formulating the equations based on AP formulas.

We use the standard formulas for an Arithmetic Progression:

(i) The $n^{th}$ term formula: $a_n = a + (n - 1)d$

(ii) The sum of $n$ terms formula: $S_n = \frac{n}{2} [a + a_n]$

Step 2: Expressing $a$ in terms of $n$ using the $n^{th}$ term formula.

Substitute the given values $a_n = 4$ and $d = 2$ into the formula $a_n = a + (n - 1)d$:

$4 = a + (n - 1)(2)$

$4 = a + 2n - 2$

$a = 4 - 2n + 2$

$a = 6 - 2n$ --- (Equation 1)

Step 3: Substituting $a$ into the sum formula.

Substitute $S_n = -14$ and $a_n = 4$ into the formula $S_n = \frac{n}{2} [a + a_n]$:

$-14 = \frac{n}{2} [a + 4]$

Multiply both sides by 2:

$-28 = n(a + 4)$ --- (Equation 2)

Step 4: Solving for $n$.

Substitute Equation 1 into Equation 2:

$-28 = n[(6 - 2n) + 4]$

$-28 = n[10 - 2n]$

$-28 = 10n - 2n^2$

Rearrange the terms to form a standard quadratic equation $ax^2 + bx + c = 0$:

$2n^2 - 10n - 28 = 0$

Divide the entire equation by 2 to simplify:

$n^2 - 5n - 14 = 0$

Factorize the quadratic equation by splitting the middle term:

$n^2 - 7n + 2n - 14 = 0$

$n(n - 7) + 2(n - 7) = 0$

$(n - 7)(n + 2) = 0$

This gives two possible values for $n$: $n = 7$ or $n = -2$.

[Since the number of terms $n$ must be a positive integer, we discard $n = -2$.]

Therefore, $n = 7$.

Step 5: Finding the first term $a$.

Substitute $n = 7$ back into Equation 1:

$a = 6 - 2(7)$

$a = 6 - 14$

$a = -8$

Final Answer:

The number of terms $n = 7$ and the first term $a = -8$.


More Questions from Class 10 Mathematics Arithmetic Progression EXERCISE 5.3


CBSE Solutions for Class 10 Mathematics Arithmetic Progression


Chapters in CBSE - Class 10 Mathematics


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