The number of terms of an A.P. is even; the sum of all the odd terms is 24, the sum of all the even terms is 30 and the last term exceeds the first by 21 2. Then the number of terms which are integers in the A.P. is:
Step-by-Step Solution
Key Concept: Use separate sums of$odd-position$and$even-position$terms of$a_n$AP to find$a,d,n$.
$a_{2} + a_{4}$+$\$ldots + an = 30$...$(1)$$(1)$$$a_{1}$+$a_{3}$$+$$\$ldots + an -1 = 24$...$(2)$$(1)$-$(2)$($a_{2} - a_{1}$) + ($a_{4} - a_{3}$)$$$\ldots ($$a_n - a_n-1$) = 6 n$$\Rightarrow$$d = 6$$\Rightarrow$$nd = 12$2 21$ an -$a_{1}$=$(n - 1)$d = 2$21 21 $\Rightarrow$$nd - d$= $\Rightarrow$ 12 - = d 2 2 3 $\Rightarrow$ d =,$n = 8$2 4 Sum of odd terms = [$2a +$(4 - 1)$3$] = 24 2 3 $\Rightarrow$$a = 2$3 9 15 21 A.P. $\Rightarrow$, 3,, 6,, 9,, 12 2 2 2 2 no. of integer$terms = 4$
Correct Answer: 1