Q63.Let the sum of an infinite G. P., whose first term is a and the common ratio is r, be 5 . Let the sum of its first five terms be 98 . Then the sum of the first 21 terms of an AP, whose first term is 10ar, nth term is an and the 25 common difference is 10 ar2 , is equal to (1) 21a11 (2) 22a11 (3) 15a16 (4) 14a16
What This Question Tests
This question tests the understanding of formulas for sum of infinite GP, sum of first n terms of GP, and sum of first n terms of AP. It requires solving for 'r' from the GP conditions and then substituting values to find the sum of the AP.
Concepts Tested
Formulas Used
S_inf = a/(1-r)
S_n_GP = a(1-r^n)/(1-r)
S_n_AP = n/2 * (2A + (n-1)D)
📚 NCERT Sections This Tests
2.1 — Two Charges 5 × 10–8 C And –3 × 10–8 C Are Located 16 Cm Apart. At
Physics Class 11 · Chapter 2
2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
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Physics Class 11 · Chapter 2
2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.
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5.12 Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
📋 Question Details
- Chapter
- Sequences & Series
- Topic
- Geometric Progression (GP) and Arithmetic Progression (AP)
- Year
- 2022
- Shift
- 27 Jul Shift 2
- Q Number
- Q63
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 9: Sequences and Series
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