Question 129

If $$a_1 = 1$$ and $$a_{n+1} = 2a_n +5$$, n=1,2,....,then $$a_{100}$$ is equal to:

Solution

$$a_2 = 2*1 + 5$$
$$a_3 = 2*(2 + 5) + 5 = 2^2 + 5*2 + 5$$
$$a_4 = 2^3 + 5*2^2 + 5*2 + 5$$
...
$$a_{100} = 2^{99} + 5*(2^{98} + 2^{97} + ... + 1)$$
$$= 2^{99} + 5*1*\dfrac{(2^{99} - 1)}{(2-1)} = 2^{99} + 5*2^{99} - 5 = 6*2^{99} - 5$$

Get 3 Free CAT Mock Tests with Expert Video Solutions Get 3 Free CAT Mock Tests

Video Solution

View More Questions From Progressions and Series

Create a FREE account and get:

All Quant CAT complete Formulas and shortcuts PDF
38+ CAT previous year papers with video solutions PDF
5000+ Topic-wise Previous year CAT Solved Questions for Free

CAT Quant Questions | CAT Quantitative Ability

CAT Geometry Questions CAT Progressions and Series Questions CAT Remainders Questions CAT Simple Interest Compound Interest Questions CAT Inequalities Questions

CAT DILR Questions | LRDI Questions For CAT

CAT Table based DI Sets Questions CAT DI Miscellaneous Questions CAT Coins and Weights Questions CAT Data change over a period Questions CAT Quant Based LR Questions

CAT Verbal Ability Questions | VARC Questions For CAT

CAT Verbal Ability Questions CAT Para Jumbles Questions CAT Grammar and Sentence Correction Questions CAT Reading Comprehension Questions CAT Odd One Out Questions

Related Formulas With Tests

Sum of common series till 'n' terms