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

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

Vedclass · Accepted Answer

$8$

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

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:

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