1. Find 10ᵗʰ term and sum of initial 10 terms of series:
3, 10, 17, 24, ……………….
Sol:
Given series is AP where a = 3 , d = 10 – 3 = 7
∴ T₁₀ = 3 + (10 – 1).7
= 66 Answer
S₁₀ = \(\frac{{40}}{2}\)[2×3 + (10 – 1).7]
= 5[69]
= 345 Answer
2. Find the 40ᵗʰ term and sum till 30ᵗʰ term of series:
32, 16, 8, 4, …………….
Sol:
Given series is a GP
where a = 32
& r = \(\frac{{13}}{{32}} = \frac{1}{2}\)
T₄₀ = 32.\({\left( {\frac{1}{2}} \right)^{40 – 1}}\) = \(\frac{1}{{{2^{34}}}}\) Answer
& S₃₀ = \(32.\frac{{\left( {1 – {{\left( {\frac{1}{2}} \right)}^{30}}} \right)}}{{1 – \frac{1}{2}}}\)
= \({2^6}\left( {1 – \frac{1}{{{2^{30}}}}} \right)\) Answer
3. How many terms of the series -16, -11, -6, ……… must be taken that the sum may be 12.
Sol:
12 = \(\frac{n}{2}\)[2×(-16) + (n – 1)(5)]
24 = n[-32 + 5n – 5]
5n² – 37n – 24 = 0
5n² – 40n + 3n – 24 = 0
(5n + 3)(n – 8) = 0
n = 8 Answer
4. How many terms are there in the series:
30, 35, 40, 45, ………….. , 875
Sol:
Number of terms = \(\frac{{875 – 30}}{5}\) + 1
= 170 terms Answer
5. A man gets Rs 100 on 1ˢᵗ april and Rs. 10 more each day than the preceding day. How much does he earn by the 18ᵗʰ may.
Sol: number of days from 1ˢᵗ april to 18ᵗʰ may = 30 + 18 = 48
So we are to find the sum of an AP whose first term is 100 and common difference is 10 & number of terms are 48
S₄₈ = \(\frac{{48}}{2}\).[2×100 + (48 – 1)10]
+ 16080 Answer
6. On 1ˢᵗ april a boy gets Rs. 1, on second day 2 rupees, on third day 4 rupees, on fourth day 8 rupees and so on till 18ᵗʰ may. How much will the boy get from 1ˢᵗ april to 18ᵗʰ may.
Sol:
This will form a GP whose
first term (a) = 1
common ratio (r) = 2
number of terms (n) = 48
∴ S₄₈ = \(\frac{{1.({2^{48}} – 1)}}{{2 – 1}}\) = (2⁴⁸ – 1) Rs. Answer
7. If the first term and last term of an aP are 93 and 407. If there are 12 terms in this series. Find the sum of the series.
Sol:
Sum of series = \(\frac{{number\;of\;terms}}{2}[first\;term + last\;term]\)
= \(\frac{{12}}{2}[93 + 407]\)
= 3000 Answer
8. The sum of an infinite GP whose common ratio is less than 1 is 32 and the sum of first two terms is 24. What will be the 5ᵗʰ term.
Sol:
\(\frac{a}{{1 – r}}\) = 32 ………………(i)
a + ar = 24
a(1 + r) = 24 ………..(ii)
divide equation (ii) with equation (i)
1 – r² = \(\frac{{24}}{{32}}\)
r² = \(\frac{1}{4}\)
r = \(\frac{1}{2}\)
put this value in equation (ii)
\(a\left( {1 + \frac{1}{2}} \right)\) = 24
a = 16
∴ Tₙ = 16×\({\left( {\frac{1}{2}} \right)^4}\) = 1 Answer
9. Find the value of \({a^{\frac{1}{3}}}.{a^{\frac{1}{9}}}.{a^{\frac{1}{{27}}}}………..\infty \)
Sol:
\(\frac{1}{3},\frac{1}{9},\frac{1}{{27}},……………\infty \) form an infinite GP whose sum will be = \(\frac{{\frac{1}{3}}}{{1 – \frac{1}{3}}}\) = \(\frac{1}{2}\)
∴ Given expression is = \({a^{\frac{1}{2}}}\) = \(\sqrt a \) Answer
10. There are 70 terms in an AP. Its 4ᵗʰ term and 67ᵗʰ term are 32 and 908 respectively. Find the sum of this AP.
Sol:
Given terms are equidistant terms from both ends respectively.
∴ Sum = \(\frac{{70}}{2}[32 + 908]\)
= 32900 Answer