Progression (Level - 3)
Q1ā10 of 50If in any decreasing arithmetic progression, sum of all its terms, except for the first term, is equal to ā36, the sum of all its terms, except for the last term, is zero, and the difference of the tenth and the sixth term is equal to ā16, then what will be first term of this series?
Step-by-step:
Let the terms be $a_1, a_2, \dots, a_n$ with common difference $d$. Since it is a decreasing AP, $d < 0$.
The difference $a_{10} - a_6 = 4d = -16 \implies d = -4$.
$S - a_1 = 36$ and $S - a_n = 0 \implies S = a_n$.
Subtracting equations: $a_1 - a_n = 36 \implies (n-1)(-d) = 36 \implies (n-1)(4) = 36 \implies n = 10$.
Total sum $S = \frac{n}{2}(a_1 + a_n) = 5(a_1 + a_n) = a_n \implies 5a_1 + 4a_n = 0$.
Substitute $a_n = a_1 - 36$: $5a_1 + 4(a_1 - 36) = 0 \implies 9a_1 = 144 \implies a_1 = 16$.
The sum of all terms of the arithmetic progression having ten terms except for the first term, is 99, and except for the sixth term, 89. Find the third term of the progression if the sum of the first and the fifth term is equal to 10.
Step-by-step:
$(S - a_6) - (S - a_1) = a_1 - a_6 = 89 - 99 = -10 \implies a_6 - a_1 = 5d = 10 \implies d = 2$.
$a_1 + a_5 = 10 \implies a_1 + (a_1 + 4d) = 10 \implies 2a_1 + 8 = 10 \implies a_1 = 1$.
Third term: $a_3 = a_1 + 2d = 1 + 2(2) = 5$.
Answer: (b) 5
Product of the fourth term and the fifth term of an arithmetic progression is 456. Division of the ninth term by the fourth term of the progression gives quotient as 11 and the remainder as 10. Find the first term of the progression.
Step-by-step:
$a_9 - 11 a_4 = 10 \implies (a_1 + 8d) - 11(a_1 + 3d) = 10 \implies -10a_1 - 25d = 10 \implies 2a_1 + 5d = -2$.
Express $d = \frac{-2 - 2a_1}{5}$.
Substitute $a_4 = a_1 + 3d = \frac{-6 - a_1}{5}$ and $a_5 = a_1 + 4d = \frac{-8 - 3a_1}{5}$ into $a_4 \cdot a_5 = 456$.
$(\frac{-6 - a_1}{5})(\frac{-8 - 3a_1}{5}) = 456 \implies 3a_1^2 + 26a_1 - 11352 = 0$.
Factoring gives $(a_1 + 66)(3a_1 - 172) = 0 \implies a_1 = -66$.
Answer: (d) -66
A number of saplings are lying at a place by the side of a straight road. These are to be planted in a straight line at a distance interval of 10 metres between two consecutive saplings. Mithilesh, the countryās greatest forester, can carry only one sapling at a time and has to move back to the original point to get the next sapling. In this manner he covers a total distance of 1.32 kms. How many saplings does he plant in the process if he ends at the starting point?
Step-by-step:
Assuming the 1st sapling is at the starting point (0 m), distance covered for 1st sapling = 0.
For the $k$-th sapling at distance $10(k-1)$ meters, distance traveled back and forth = $20(k-1)$.
Total distance = $\sum_{k=1}^n 20(k-1) = 20 \times \frac{(n-1)n}{2} = 10n(n-1) = 1320 \text{ m}$.
$n(n-1) = 132 \implies n = 12$.
Answer: (d) 12
A geometric progression consists of 500 terms. Sum of the terms occupying the odd places is Pā and the sum of the terms occupying the even places is Pā. Find the common ratio.
Step-by-step:
$P_1 = a + ar^2 + ar^4 + \dots + ar^{498} = a \frac{1 - r^{500}}{1 - r^2}$.
$P_2 = ar + ar^3 + ar^5 + \dots + ar^{499} = r \left(a + ar^2 + \dots + ar^{498}\right) = r P_1$.
Common ratio $r = \frac{P_2}{P_1}$.
Answer: (a) $P_2 / P_1$
The sum of the first ten terms of the geometric progression is Sā and the sum of the next ten terms (11th through 20th) is Sā. Find the common ratio.
Ā (a) (Sā/Sā)1/10
(b)Ā ā (Sā/Sā)1/10
(c) Ā±Ā (ā(Sā/Sā))1/10
(d) (Sā/Sā)1/5
Step-by-step:
$S_1 = a + ar + \dots + ar^9 = \frac{a(r^{10} - 1)}{r - 1}$.
$S_2 = ar^{10} + ar^{11} + \dots + ar^{19} = r^{10} \frac{a(r^{10} - 1)}{r - 1} = r^{10} S_1$.
$r^{10} = \frac{S_2}{S_1} \implies r = \pm \sqrt[10]{\frac{S_2}{S_1}}$.
Answer: (c) $\pm \sqrt[10]{\frac{S_2}{S_1}}$
The first and the third terms of an arithmetic progression are equal, respectively, to the first and the third term of a geometric progression, and the second term of the arithmetic progression exceeds the second term of the geometric progression by 0.25. Calculate the sum of the first five terms of the arithmetic progression if its first term is equal to 2.
tep-by-step:
AP terms: $2, 2+d, 2+2d \implies a_3 = 2+2d$.
GP terms: $2, g_2, 2+2d \implies g_2 = \sqrt{2(2+2d)} = 2\sqrt{1+d}$.
$a_2 = g_2 + 0.25 \implies 2 + d = 2\sqrt{1+d} + 0.25 \implies d + 1.75 = 2\sqrt{1+d}$.
Squaring both sides: $(d + 1.75)^2 = 4(1+d) \implies d = 1.25$ or $d = -0.75$.
For $d = -0.75$, $S_5 = \frac{5}{2}(2(2) + 4(-0.75)) = 2.5$.
If (2 + 4 + 6 + ā¦ 50 terms)/(1 + 3 + 5 + ā¦ n terms) = 51/2, then find the value of n.
Step-by-step:
Numerator $= 2(1 + 2 + \dots + 50) = 50 \times 51$.
Denominator $= 1 + 3 + 5 + \dots + (2n-1) = n^2$.
$\frac{50 \times 51}{n^2} = \frac{51}{2} \implies \frac{50}{n^2} = \frac{1}{2} \implies n^2 = 100 \implies n = 10$.
Answer: (d) 10
\( (\,\underbrace{666\ldots}_{n\ \text{digits}}\,)^2+(\,\underbrace{888\ldots}_{n\ \text{digits}}\,) \)
is equal to
(a) \( \dfrac{4}{9}(10^n-1) \)
(b) \( \dfrac{4}{9}(10^{2n}-1) \)
(c) \( \dfrac{4(10^n-10^{n-1}-1)}{9} \)
(d) \( \dfrac{4(10^n+1)}{9} \)
Step-by-step:
$666\dots n\text{ digits} = \frac{6(10^n - 1)}{9} = \frac{2}{3}(10^n - 1)$.
Square $= \frac{4}{9}(10^n - 1)^2$.
$888\dots n\text{ digits} = \frac{8}{9}(10^n - 1)$.
Sum $= \frac{4}{9}(10^n - 1)\left[(10^n - 1) + 2\right] = \frac{4}{9}(10^n - 1)(10^n + 1) = \frac{4}{9}(10^{2n} - 1)$.
Answer: (b) $(10^{2n} - 1) \times \frac{4}{9}$
The interior angles of a polygon are in AP. The smallest angle is 120Ā° and the common difference is 5Ā°. Find the number of sides of the polygon.
Step-by-step:
Sum of interior angles of $n$-gon $= (n-2) \times 180^\circ$.
AP sum $= \frac{n}{2}\left(2(120) + (n-1)5\right) = (n-2) \times 180$.
$5n^2 - 125n + 720 = 0 \implies n^2 - 25n + 144 = 0 \implies (n-9)(n-16) = 0$.
If $n = 16$, max angle $= 120^\circ + 15(5^\circ) = 195^\circ > 180^\circ$ (impossible for convex polygon). Thus $n = 9$.
Answer: (c) 9