With respect to the usual notations of an A.P | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: $T_{n} = a + (n - 1)d = 4$ and $S_{n} = \frac{n}{2}[a + T_{n}] = -14$. Substituting $T_{n} = 4$ in the sum formula: $\frac{n}{2}[a + 4] = -

Vedclass · Accepted Answer

$n = 7, a = -8$

Vedclass · Answer

(A) Given: $T_{n} = a + (n - 1)d = 4$ and $S_{n} = \frac{n}{2}[a + T_{n}] = -14$. Substituting $T_{n} = 4$ in the sum formula: $\frac{n}{2}[a + 4] = -14 \implies n(a + 4) = -28$. From the first equation: $a = 4 - (n - 1)d = 4 - 2n + 2 = 6 - 2n$. Substitute $a$ in the sum equation: $n(6 - 2n + 4) = -28 \implies n(10 - 2n) = -28$. $10n - 2n^{2} = -28 \implies 2n^{2} - 10n - 28 = 0 \implies n^{2} - 5n - 14 = 0$. Factoring the quadratic: $(n - 7)(n + 2) = 0$. Since $n$ must be a positive integer,$n = 7$. Now,find $a$: $a = 6 - 2(7) = 6 - 14 = -8$. Thus,$n = 7$ and $a = -8$.

With respect to the usual notations of an $A.P.$,if $T_{n} = 4$,$d = 2$,and $S_{n} = -14$,find $n$ and $a$.

Similar Questions

Find the sum of the integers between $100$ and $200$ that are not divisible by $9$.

If $2, x, 20$ are three consecutive terms of an $A.P.$,then $x = ..........$

For an $A.P.$,$a=11$ and $d=7$. Find the sum of the first $40$ terms of the $A.P.$

The first term of an $AP$ is $-5$ and the last term is $45$. If the sum of the terms of the $AP$ is $120$,then find the number of terms and the common difference.

Find the sum of first $n$ terms of the $A.P.$ $6, 2, -2, \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module