The number of terms in an A.P. is even. The s | vedclass

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

Question

(B) Let the total number of terms be $2n$,the first term be $a$,and the common difference be $d$.The terms are $a, a+d, a+2d, ..., a+(2n-1)d$.The odd-

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let the total number of terms be $2n$,the first term be $a$,and the common difference be $d$.The terms are $a, a+d, a+2d, ..., a+(2n-1)d$.The odd-positioned terms are $a, a+2d, ..., a+(2n-2)d$. There are $n$ such terms.Sum of odd terms $S_o = \frac{n}{2}[2a + (n-1)(2d)] = n[a + (n-1)d] = 24$  --- $(i)$The even-positioned terms are $a+d, a+3d, ..., a+(2n-1)d$. There are $n$ such terms.Sum of even terms $S_e = \frac{n}{2}[2(a+d) + (n-1)(2d)] = n[a+d+(n-1)d] = 30$  --- $(ii)$Subtracting $(i)$ from $(ii)$:$n(a+d+(n-1)d) - n(a+(n-1)d) = 30 - 24$$nd = 6$  --- $(iii)$Given the last term exceeds the first term by $10\frac{1}{2} = \frac{21}{2}$:$(a+(2n-1)d) - a = \frac{21}{2}$$(2n-1)d = \frac{21}{2}$$2nd - d = \frac{21}{2}$Substitute $nd = 6$ into the equation:$2(6) - d = \frac{21}{2}$$12 - d = 10.5$$d = 1.5 = \frac{3}{2}$Using $nd = 6$:$n(\frac{3}{2}) = 6 \Rightarrow n = 4$Total number of terms $= 2n = 2 	imes 4 = 8$.

The number of terms in an $A.P.$ is even. The sum of the odd terms is $24$ and the sum of the even terms is $30$. If the last term exceeds the first term by $10\frac{1}{2}$,then the number of terms in the $A.P.$ is:

Similar Questions

Let $a_1=8, a_2, a_3, \ldots, a_n$ be an $A.P.$ If the sum of its first four terms is $50$ and the sum of its last four terms is $170$,then the product of its middle two terms is

The value of $\sum\limits_{r = 1}^n {\log \left( {\frac{{{a^r}}}{{{b^{r - 1}}}}} \right)} $ is

If the ratio of the sum of $n$ terms of two arithmetic progressions is $2n + 3 : 6n + 5$,then the ratio of their $13^{th}$ terms is.......

Let $A_1, A_2, \dots, A_{39}$ be $39$ arithmetic means between the numbers $59$ and $159$. Then the mean of $A_{25}, A_{28}, A_{31}$ and $A_{36}$ is equal to:

Let $A, G, H$ and $S$ respectively denote the arithmetic mean,geometric mean,harmonic mean and the sum of the numbers $a_1, a_2, a_3, \ldots, a_n$. Then the value of $x$ at which the function $f(x)=\sum_{k=1}^n(x-a_k)^2$ has a minimum is

Vedclass Test Series

Exam Paper Generator

Online Exam Module