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8. Sequences and Series
easy
શ્રેણી $2, 5, 8, 11,…..$ ના $n$ પદોનો સરવાળો $60100$ હોય, તો $n = …..$
A
$100$
B
$150$
C
$200$
D
$250$
Solution
$a = 2, d = 3, S_n = 60100$
${S_n} = \frac{n}{2}[2a + (n – 1)d]\,$
$\therefore \,\,\frac{n}{2}[4 + (n – 1)3] = 60100\,\,\,\,\,$
$\therefore \,\,\,n(3n + 1) = 120200$ $\therefore \,\,3{n^2} + n – 120200 = 0\,\,\,$
$\therefore \,3{n^2} – 600n + 601n – 120200 = 0$
$(3n + 601)(n – 200) = 0$ અને $n > 0$ $n = 200$
Standard 11
Mathematics