an=a1+(n−1)d
300=1+(n−1)d
⇒d=(n−1)299=(n−1)13×23= integer
so n−1=±13,±23,±299,±1
⇒n=14,−12,24,−22,300,−298,2,0
But n∈[15,50]⇒n=24⇒d=13
Hence,
Sn−4=S20=220[2(1)+(20−1)(13)]
⇒Sn−4=2490
And,
an−4=a20=a1+19d
=1+19×13
=248
Let a,1a2,…,an be a given A.P. whose common difference is an integer and Sn=a1+a2+…+an. If a1=1,an=300 and 15≤n≤50, then the ordered pair (Sn−4,an−4) is equal to:
Held on 4 Sept 2020 · Verified 6 Jul 2026.
(2490,249)
(2480,249)
(2480,248)
(2490,248)
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
If the roots of x² - 5x + k = 0 are in the ratio 2:3, then k equals:
Let $\alpha = 3+4+8+9+13+14+\ldots$ upto 40 terms. If $(\tan\beta)^{\frac{\alpha}{1020}}$ is a root of the equation $x^2+x-2=0$, $\beta \in \left(0, \dfrac{\pi}{2}\right)$, then $\sin^2\beta + 3\cos^2\beta$ is equal to:
If the set of all solutions of $|x^2 + x - 9| = |x| + |x^2 - 9|$ is $[\alpha, \beta] \cup [\gamma, \infty)$, then $(\alpha^2 + \beta^2 + \gamma^2)$ is equal to:
The sum of squares of all the real solutions of the equation $\log_{(x+1)}(2x^2+5x+3) = 4 - \log_{(2x+3)}(x^2+2x+1)$ is equal to ________.
Let $f:(1,\infty)\to\mathbb{R}$ be a function defined as $f(x) = \dfrac{x-1}{x+1}$. Let $f^{i+1}(x) = f(f^i(x))$, $i=1, 2, \ldots, 25$, where $f^1(x)=f(x)$. If $g(x) + f^{26}(x) = 0$, $x \in (1, \infty)$, then the area of the region bounded by the curves $y=g(x)$, $2y=2x-3$, $y=0$ and $x=4$ is:
Work through every JEE Main Algebra PYQ, year by year.