1
JEE Main 2019 (Online) 10th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
The sum
$$1 + {{{1^3} + {2^3}} \over {1 + 2}} + {{{1^3} + {2^3} + {3^3}} \over {1 + 2 + 3}} + ...... + {{{1^3} + {2^3} + {3^3} + ... + {{15}^3}} \over {1 + 2 + 3 + ... + 15}}$$$$ - {1 \over 2}\left( {1 + 2 + 3 + ... + 15} \right)$$ is equal to :
A
620
B
1240
C
1860
D
660
2
JEE Main 2019 (Online) 10th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Let $$a$$, b and c be in G.P. with common ratio r, where $$a$$ $$ \ne $$ 0 and 0 < r $$ \le $$ $${1 \over 2}$$ . If 3$$a$$, 7b and 15c are the first three terms of an A.P., then the 4th term of this A.P. is :
A
$$a$$
B
$${7 \over 3}a$$
C
5$$a$$
D
$${2 \over 3}a$$
3
JEE Main 2019 (Online) 10th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Let a1, a2, a3,......be an A.P. with a6 = 2. Then the common difference of this A.P., which maximises the product a1a4a5, is :
A
$${3 \over 2}$$
B
$${6 \over 5}$$
C
$${8 \over 5}$$
D
$${2 \over 3}$$
4
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
The sum
$${{3 \times {1^3}} \over {{1^3}}} + {{5 \times ({1^3} + {2^3})} \over {{1^2} + {2^2}}} + {{7 \times \left( {{1^3} + {2^3} + {3^3}} \right)} \over {{1^2} + {2^2} + {3^2}}} + .....$$ upto 10 terms is:
A
600
B
660
C
680
D
620
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