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

How many terms of the $AP: -15, -13, -11, \ldots$ are needed to make the sum $-55$? Explain the reason for the double answer.

Find four numbers in $A.P.$ such that their sum is $36$ and the product of means ($2^{nd}$ and $3^{rd}$ term) exceeds the product of extremes ($1^{st}$ and $4^{th}$ term) by $32.$

For an $A.P.$,the $12^{th}$ term is $-13$ and the sum of its first $4$ terms is $24$. Find the sum of the first $10$ terms of the $A.P.$

For an $A.P.$,the first term is $10$,the $n^{th}$ term is $40$,and the sum of the first $n$ terms is $150$. Find the value of $n$ and the common difference $d$.

Which term of the $A.P.$ $40, 36, 32, \ldots$ is $0$? Also,find the sum of how many terms of this $A.P.$ is $0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module