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Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000?

\\$As per the question she saves 32 in first, 36 in second and as so on.\\ 32, 36, 40, $ \ldots $ $ \ldots $ \\$ $ Here, first term (a) = 32\\ Common difference (d) = 36 - 32 = 4\\ Let in n month she saves Rs. 2000\\$ {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\ 2000=\frac{\mathrm{n}}{2}[2 \times 32+(\mathrm{n}-1) 4] 4000 = n[64 + 4n - 4]\\ 4000 = n[60 + 4n]\\ 4n\textsuperscript{2} + 60n - 4000 = 0\\ Divide by 4\\ n\textsuperscript{2} + 15n - 1000 = 0\\ n\textsuperscript{2} + 40n - 25n - 1000 = 0\\ n(n + 40) - 25 (n + 40) = 0\\ (n + 40) (n - 25) = 0\\ n = - 40$(months cannot be negative, which is not possible)$, n = +25\\ $Hence it takes 25 months to save Rs. 2000.\\ $

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Kanika was given her pocket money on Jan 1st, 2008. She puts Re 1 on Day 1,Rs 2 on Day 2, Rs 3 on Day 3, and continued doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money, and found that at the end of the month she still had Rs 100 with her. How much was her pocket money for the month?

According to the question she puts Rs. 1 on Day 1, 2 on Day 2 and so on. 

\\$Hence 1, 2, 3,$ \ldots $ $ \ldots $ .. , 31\\ where a $= 1\\ d = 2 - 1 = 1 \\ a\textsubscript{n} = 31\\ $Total money she put, $ {{S}_{n}}=\frac{n}{2}\left[ a+{{a}_{n}} \right] \\

\\ \mathrm{S}_{\mathrm{n}}=\frac{31}{2}[1+31] \\\\ \mathrm{S}_{\mathrm{n}}=\frac{31}{2} \times 32 \Rightarrow 31 \times 16=496

Total money = Total she put + Spent + left 

= 496 + 204 + 100 

Total = 800 Rs 

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The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another AP whose first term is - 30 and the common difference is 8. Find n.

\\ $Given: a = 8, d\textsubscript{1} $= 20 for first AP\\ b = - 30, d\textsubscript{2} = 8 $for second AP\\$ $ Given $ S\textsubscript{n} = S\textsubscript{2n}\\ \frac{\mathrm{n}}{2}\left[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}_{1}\right]=\frac{2 \mathrm{n}}{2}\left[2 \mathrm{~b}+(2 \mathrm{n}-1) \mathrm{d}_{2}\right] \\ \frac{\mathrm{n}}{2}[2(8)+(\mathrm{n}-1) 20]=\mathrm{n}[2(-30)+(2 \mathrm{n}-1)(8)] \\ \frac{\mathrm{n}}{2}[16+20 \mathrm{n}-20]=\mathrm{n}[-60+16 \mathrm{n}-8] \\ \frac{\mathrm{n}}{2}[20 \mathrm{n}-4]=\mathrm{n}[-68+16 \mathrm{n}] \\ \frac{2(10 \mathrm{n}-2)}{2}=-68+16 \mathrm{n} 10n - 16n = - 68 + 2\\ -6n = - 66\\ n = \frac{+66}{+6}=11 \\ n = 11\\

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How many terms of the AP: -15, -13, -11, _______ are needed to make the sum -55? Explain the reason for double answer.

Here the given AP is -15, -13, -11$ \ldots $ \\

 First term (a) = - 15 

 Common difference (d) = - 13 - (-15) = - 13 + 15 = 2 

 Let n terms have sum - 55 then 

{{S}_{n}}=-55

{{S}_{n}} =\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]

\\ n[-30 + 2n - 2] = - 55 $ \times $ 2\\ -32n + 2n\textsuperscript{2} + 110 = 0\\ 2n\textsuperscript{2} - 32n + 110 = 0\\ n\textsuperscript{2} - 16n + 55 = 0\\ n\textsuperscript{2} - 5n - 11n + 55 = 0\\ n(n - 5) - 11(n - 5) = 0\\ (n - 5) (n - 11) = 0\\ n = 5, 11\\

There are two answers which are 5th when n=5 all terms are negative and 11th terms when n=11  the AP will contain positive numbers and negative terms so the resulting sum is - 55.  

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Find the sum of first seven numbers which are multiples of 2 as well as of 9.

\\$LCM of 2 and 9 is 18.\\ So the terms are 18, 36, $ \ldots $ .., 126\\ First term (a) = 18\\

$Common difference (d) = 36 - 18 = 18\\

{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]

\\ $Put n=7\\

\\ =\frac{7}{2}\left[ 36+108 \right] \\ =\frac{7}{2}\left[ 144 \right] \\

\\S\textsubscript{7} = 7 $ \times $ 72 \\ S\textsubscript{7} = 504\\

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Find the sum of last ten terms of the AP: 8, 10, 12, __________ , 126.

\\$The given AP is 8, 10, $ \ldots $ .., 126\\ AP from last is 126, 124, $ \ldots $ ., 10, 8\\ a = 126\\ d = 124 - 126 = - 2\\

Sum of 10 terms from last-
$$ S_{n}=\frac{n}{2}[2 a+(n-1) d] $$
$Put $n=10$
$$ \begin{aligned} \mathrm{S}_{10} &=\frac{10}{2}[2(126)+(10-1)(-2)] \\ &=5[252+9(-2)] \\ &=5[252-18] \\ \mathrm{S}_{10} &=5 \times 234=1170 \\ \mathrm{~S}_{10} &=1170 \end{aligned} $$

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Find the sum of all the 11 terms of an AP whose middlemost term is 30.

\\$The middle term of an AP$ = \frac{n+1}{2} \\ $Here n $= 11 \\ $Hence a\textsubscript{6}$ = 30\\ a + 5d = 30 $ \ldots $ (1) (Using a\textsubscript{n} = a + (n - 1)d)\\ {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] put n=11\\ =\frac{11}{2}\left[ 2a+10d \right] \\ =\frac{11}{2}\left[ 2\left( a+5d \right) \right] \\ = 11 [30] (Using (1))\\ S\textsubscript{11} = 330\\

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If sum of first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

\\S\textsubscript{6} = 36, S\textsubscript{16} = 256\\ \left[ Using\,\,{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \right] \\ S\textsubscript{6} = 36 S\textsubscript{16} = 256\\ = \frac{6}{2}\left[ 2a+\left( 6-1 \right)d \right]=36 \frac{16}{2}\left[ 2a+\left( 16-1 \right)d \right]=256 \\ =3[2a + 5d] = 36 8[2a + 15d] = 256\\ =2a + 5d = 12 $ \ldots $ (1) \\2a + 15d = 32 $ \ldots $ (2)\\\text{from (1)and (2)}

\\ \_\_\_\_\_\_\_\_\_\_\_\_\\ 10d = 20\\ d = 2\\ Put \ d = 2\ in (1)\\ 2a + 5(2) = 12\\ 2a = 12 - 10\\ a = 1\\ ( \left[ Using\,\,{{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \right] \\ Put n= 1\\ = 5 [2 + 18] \\ 5[20]\\ S\textsubscript{10} = 100\\

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Find the sum of first 17 terms of an AP whose 4th and 9th terms are -15 and -30 respectively.

\\$a\textsubscript{4 }= - 15, a\textsubscript{9} = - 30\\ a\textsubscript{4} = - 15 a\textsubscript{9} = - 30\\ a + 3d = - 15 $ \ldots $ (1) a + 8d = - 30 $ \ldots $ (2) (Using a\textsubscript{n} = a + (n - 1).d)\\ from equation (1) $\&$ (2) \\

\\-5d = 15\\ d = - 3\\ $ Put d = - 3 in (1)\\$ a + 3(-3) = - 15\\ a = - 15 + 9\\ a = - 6\\ {{S}_{17}}=\frac{17}{2}\left[ 2\left( -6 \right)+\left( 17-1 \right)\left( -3 \right) \right] \\Using \ {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\ \begin{aligned} & =\frac{17}{2}\left[ -12-48 \right] \ & =\frac{17}{2}\left[ -60 \right] \ \end{aligned} \\ S\textsubscript{17} = - 510\\

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If S\textsubscript{n} denotes the sum of first n terms of an AP, prove that \\ S\textsubscript{12} = 3(S\textsubscript{8} - S\textsubscript{4})\\

\\\text{To prove}:- S\textsubscript{12} = 3(S\textsubscript{8} - S\textsubscript{4})\\ If \ S\textsubscript{n} \text{is the sum of n terms then first term is a and common difference is d} \\ S_{n}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \ldots \ldots .. (1)\\ \text{Put }n = 12 in equation (1) \\ S\textsubscript{12} = 6[2a + 11d]\\ S\textsubscript{12} = 12a + 66d \ldots \ldots \ldots \ldots . (1)\\ S\textsubscript{8} = 4[2a + 7d]\\ S\textsubscript{8} = 8a + 28d \ldots \ldots \ldots \ldots \ldots .(2)\\ S\textsubscript{4} = 2[2a + 3d]\\ S\textsubscript{4} = 4a + 6d \ldots \ldots \ldots \ldots \ldots ..(3)\\ RHS \\3(S\textsubscript{8} - S\textsubscript{4}) = \\ =3(8a + 28d - 4a - 6d) \text{By Equation (2) and (3)}\\ =3(4a + 22d)\\ = 12a + 66d By Equation (1)\\ = S\textsubscript{12}\\ \text{Hence } \textsubscript{12} = 3(S\textsubscript{8} - S\textsubscript{4})

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