Arithmetic Progression
Category : 10th Class
Arithmetic Progressions
Arithmetic Progression
e.g. (i) A rule defined as \[{{T}_{n}}\text{ }=\text{ }5n\text{ }+\text{ }1\] gives
\[{{T}_{1}}=\text{ }6,\text{ }{{T}_{2}}=\text{ }11,\text{ }{{T}_{3}}=\text{ }16,\text{ }{{T}_{4}}=\text{ }21\], .............
Thus, the numbers \[6,\text{ }11,\text{ }16,\text{ }21..........\] from a sequence.
The general form of an AP is; \[a,\text{ }a\text{ }+\text{ }d,\text{ }a\text{ }+\text{ }2d,\text{ }a\text{ }+\text{ }3d...\]
we get the arithmetic series \[5\text{ }+\text{ }8\text{ }+\text{ }11\text{ }+\text{ }14\text{ }+\text{ }17..........\]
APIs given by: \[S=\frac{n}{2}(a+l)\,\,\,\,\,or\,\,S=\frac{n}{2}[2a+(n-1)d]\]
Snap Test
(a) 37 (b) 35
(c) 40 (d) 30
(e) None of these
Ans. (a)
Explanation: In the given sequence, \[a\text{ }=\text{ }3\text{ }and\text{ }d\text{ }=\text{ }6\text{ }-\text{ }3\text{ }=\text{ }3.\]
Let there be n terms in the given sequence, then \[{{t}_{n}}=\text{ }111\].
\[\therefore \] \[a\text{ }+\text{ }\left( n\text{ }-\text{ }1 \right)\text{ }d\text{ }=\text{ }111\]
\[\Rightarrow \,\,\,\,3+\left( n-1 \right)~\,\times \,\,3=111\,\,\,\Rightarrow \,\,~3\text{ }\left( n\text{ }-\text{ }1 \right)=108\,\,\,\Rightarrow \,\,~n\text{ }-\text{ }1\text{ }=\text{ }36\,\,\,\Rightarrow \,\,~n\text{ }=\text{ }37.\]
(a) \[5,\text{ }10,\text{ }15,\text{ }20,\text{ }25..........~\]
(b) \[8,\text{ }10,\text{ }12,\text{ }14,\text{ }16........\]
(c) \[8,\text{ }10,\text{ }13,\text{ }17,\text{ }22.........~\]
(d) \[5,\text{ }10,\text{ }10,\text{ }14,\text{ }12\ldots \ldots .\]
(e) None of these
Ans. (b)
Explanation: Let the first term of the A.P. be a and its common difference be d
We know that; \[{{t}_{n}}=\text{ }a\text{ }+\text{ }\left( n\text{ }-\text{ }1 \right)\text{ }d\]
Now, \[{{t}_{7}}=\text{ }20~~~~~\Rightarrow ~~~~a\text{ }+\text{ }6d\text{ }=\text{ }20\] …... (i)
And \[{{t}_{13}}=\text{ }32~~~~~\Rightarrow ~~~~~a\text{ }+\text{ }12d\text{ }=\text{ }32\] ….... (ii)
Subtracting (i) from (ii) we get: \[6d\text{ }=\text{ }12\,\,\,\,\Rightarrow \,\,\,~d\text{ }=\text{ }2.\]
Substituting \[d\text{ }=\text{ }2\] in (i) we get: \[a\text{ }+\text{ }12\text{ }=\text{ }20~\,\,\,\Rightarrow \,\,\,a\text{ }=\text{ }8.\]
(a) \[3,\text{ }7,\text{ }8,\text{ }10.........\]
(b) \[2,\text{ }7,\text{ }11,\text{ }12.........\]
(c) \[3,\text{ }7,\text{ }11,\text{ }15........\]
(d) \[3,\text{ }7,\text{ }10,\text{ }12.........\]
(e) None of these
Ans. (c)
Explanation: Let the first term be a and the common difference be d.
Then \[{{t}_{8}}=\text{ }31~\,\,\,\,\Rightarrow \,\,\,a\text{ }+\text{ }7d\text{ }=\text{ }31\] …... (i)
And \[{{t}_{15}}=\text{ }16\text{ }+\text{ }{{t}_{11}}~\,\,\Rightarrow \,\,\,a\text{ }+\text{ }14d\text{ }=\text{ }16\text{ }+\text{ }\left( a\text{ }+\text{ }10d \right)\]
\[\Rightarrow \,\,\,\,4d\text{ }=\text{ }16\,\,\,\Rightarrow \,\,\,\,~d\text{ }=\text{ }4\]
Substituting \[d\text{ }=\text{ }4\] in (i) we get: \[a+28\text{ }=\text{ }31~\,\,\,\Rightarrow \,\,a\text{ }=\text{ }3.\]
Thus, \[a\text{ }=\text{ }3\text{ }and\text{ }d\text{ }=\text{ }4\]
Hence, the required A.P. is \[3,\text{ }7,\text{ }11,\text{ }15,...\]
(a) 16 seconds (b) 18 seconds
(c) 12 seconds (d) 20 seconds
(e) None of these
Ans. (a)
Explanation: Distance covered by the body in\[1st\text{ }second\text{ }=\text{ }16\text{ }metres\],
Distance covered by the body in\[2nd\text{ }second\text{ }=\text{ }48\text{ }metres\],
Distance covered by the body in 3rd second = 80 metres and so on.
These distances form an A.P. \[16,\,\,48,\text{ }80,\text{ }.....\] having first term \[a\text{ }=\text{ }16\]and common difference \[d\text{ }=\text{ }48\text{ }-\text{ }16\text{ }=\text{ }32\].
Let the body covers the distances 4096 metres in n seconds.
Then, \[{{S}_{n}}=\text{ }4096\] \[\Rightarrow \] \[\frac{n}{2}\,\,\times \,\,[2a\,+\,\,(n\,-\,\,1)\,d]\,\,=\,\,4096\]
\[\Rightarrow \] \[\frac{n}{2}\,\,\times \,\,[(2\,\,\times \,\,16)\,\,+\,\{\,(n\,\,-\,\,1)\,\,\times \,\,32\}]\,\,=\,\,4096\]
\[\Rightarrow \] \[~\frac{n}{2}\,\,\times \left( 32n \right)=4096\text{ }\Rightarrow \text{ }16{{n}^{2}}=\text{ }4096~\,\,\Rightarrow \,\,{{n}^{2}}\,\,=\text{ }256\]
\[\Rightarrow \] \[n\,\,=\,\,16\].
Thus, the body falls through 4096 m in 16 seconds.
(a) Rs. 5.50, Rs. 1085 (b) Rs. 8.50, Rs. 1185
(c) Rs. 9.50, Rs. 1185 (d) Rs. 9.50, Rs. 1200
(e) None of these
Ans. (c)
Explanation: The charges for boring the 1st metre of the well = Rs. 5;
The charges for boring the 2nd metre of the well = Rs. \[[5+(0.50\times ~1)]\text{ }=\text{ }Rs.\text{ }5.50;\]
The charges for boring the 3rd metre of the well = Rs. \[[5+(0.50~\,\times \,~2)]\text{ }=\text{ }Rs.\text{ }6\];
………….. ………….. ………… ………….
The charges for boring the 10th metre of well = Rs. \[[5+(0.50\,~\times \,9)]\text{ }=\text{ }Rs.\text{ }9.50.\]
Thus, the charges for boring each metre form an A.P. having first term \[a\text{ }=\text{ }5\]and common difference\[d\text{ }=\text{ }0.50\].
\[\therefore \] Total cost of boring a well n metres deep, \[{{S}_{n}}\,\,=\,\,\frac{n}{2}\,\,[2a+(n-1)d]\]and so, the total cost of boring a well 60 m deep,
S60 = Rs. \[\left[ \frac{60}{2}\times \,\,\{2\,\,\times \,\,5\}\,\,+\,\,(59\,\,\times \,\,0.50)\} \right]\]
= Rs. \[[30\times \left\{ 10\,\,+\,\,29.50 \right\}]\]
\[=Rs.\text{ }(30\,\,\times \,\,39.5)\text{ }=\text{ }Rs.\text{ }1185.\]
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