Sequences and Series Test - 67

As we know,

Sum of the first \(n\) Natural Numbers:

\(1+2+3+4+\ldots+n=\sum n=\frac{ n ( n +1)}{2}\)

Sum of the square of the first \(n\) Natural Numbers:

\(1^{2}+2^{2}+3^{2}+4^{2}+\ldots+n^{2}=\sum n^{2}=\frac{n(n+1)(2 n+1)}{6}\)

Sum of the cubes of the first \(n\) Natural Numbers:

\(1^{3}+2^{3}+3^{3}+4^{3}+\ldots+ n ^{3}=\sum n ^{3}=\frac{[ n ( n +1)]^{2}}{4}\)

Given,

Series \(1^{2}+\left(1^{2}+2^{2}\right)+\left(1^{2}+2^{2}+3^{2}\right)+\ldots . .+\) up to \(n\) terms.

\(n^{th}\) term of the given series is \(T_{n}=1^{2}+2^{2}+3^{2}+\ldots . .+n^{2}\).

So, \(T_{n}=\sum n ^{2}=\frac{ n ( n +1)(2 n +1)}{6}\)

\(=\frac{1}{6}\left[2 n ^{3}+3 n ^{2}+ n \right]\)

Sum of \(n\) terms of the series, \(S_{ n }=\sum T _{ n }\)

\(\Rightarrow S _{ n }=\frac{1}{6} \sum\left(2 n ^{3}+3 n ^{2}+ n \right)\)

\(\Rightarrow S _{ n }=\frac{1}{3} \sum n ^{3}+\frac{1}{2} \sum n ^{2}+\frac{1}{6} \sum n\)

\(\Rightarrow S _{ n }=\frac{[ n ( n +1)]^{2}}{12}+\frac{ n ( n +1)(2 n +1)}{12}+\frac{ n ( n +1)}{12} \)

\(\Rightarrow S _{ n }=\frac{ n ( n +1)^{2}( n +2)}{12}\)

As we have to find sum of 12 terms of series:

So, \(n=12\)

\(\therefore S _{12}=\frac{12(12+1)^{2}(12+2)}{12}\)

\(\Rightarrow S _{12}=\frac{12 \times (13)^{2}\times (14)}{12}\)

\(\Rightarrow S _{ 12}=169 \times 14\)

\(\Rightarrow S _{ 12}=2366\)

Given,

\(a x^{2}+2 b x+c=0\)

\(\Rightarrow a x^{2}+2 \sqrt{a c} x+c=0 \quad\left(\because b^{2}=a c\right) \)

\(\Rightarrow(x \sqrt{a}+\sqrt{c})^{2}=0 \)

\(\Rightarrow x=-\frac{\sqrt{c}}{\sqrt{a}}\)

This satisfies \(dx ^{2}+2 ex + f =0\)

\(\Rightarrow d \left[\frac{ c }{ a }\right]+2 e \left[-\frac{\sqrt{ c }}{\sqrt{ a }}\right]+ f =0 \)

\(\Rightarrow\left(\frac{ dc }{ a }+ f \right)=2 e \sqrt{\frac{ c }{ a }} \)

\(\Rightarrow\left(\frac{ d }{ a }+\frac{ f }{ c }\right)=2 e \sqrt{\frac{1}{ ac }} \)

\(\Rightarrow \frac{ d }{ a }+\frac{ f }{ c }=\frac{2 e }{ b }\)

Therefore, \(\frac{ d }{ a }, \frac{ e }{ b }, \frac{ f }{ c }\) are in AP.

Numbers divisible by \(6\) between \(100\) to \(400\) are \(102, 108, 114...396\) forming an AP series.

First term \((a)=102\)

Common difference \((d)=108-102=6\)

\(n ^{\text {th }}\) term \((a_n)=396\)

As we know,

\(n^{\text {th }}\) term of AP is given by,

\(a_{n}=a+(n-1) d\)

\(\Rightarrow 396=102+(n-1) \times 6 \)

\(\Rightarrow 396-102=(n-1) \times 6 \)

\(\Rightarrow 294=(n-1) \times 6 \)

\(\Rightarrow \frac{294}{6}=(n-1)\)

\(\Rightarrow 49=n-1 \)

\(\therefore n=50\)

Now,

Sum of all series \((S_n)=\frac{ n }{2}(a+l)\) (where \(l=\) last term)

\(\therefore\) Sum of all numbers \(=\frac{50}{2}(102+396)\)

\(=25 \times 498\)

\(=12450\)

So, the sum of all numbers divisible by \(6\) in between \(100\) to \(400\) is \(12450\).

Given,

AP series \(=\frac{2}{3}, K , \frac{5}{8}\)

Let,

\(a =\) first term \( =\frac{2}{3}\)

\(b =\) second term \(= K\)

\(c=\) third term \(=\frac{5}{8}\)

Since, given three terms are in AP.

So, \(d = b - a = c - b \) (where \(d =\) common difference\()\)

\(\therefore K -\frac{2}{3}=\frac{5}{8}- K\)

\(\Rightarrow 2 K =\frac{2}{3}+\frac{5}{8}\)

\(\Rightarrow 2 K =\frac{15+16}{24}\)

\(\Rightarrow 2 K =\frac{31}{24}\)

\(\Rightarrow K =\frac{31}{48}\)

So, the value of \(K\) is \(\frac{31}{48}\).

Let the three terms be \(\frac{a}{r}, a\) and \(ar\).

Where \(a\) and \(r\) are the first term and common ratio of the GP.

Given,

The product of three terms in a GP \(=27\)

\(\therefore \frac{ a }{ r } \times a \times ar =27 \)

\(\Rightarrow a ^{3}=27\)

\(\Rightarrow a ^{3}=3^3\)

\(\therefore a=3\)

So, the middle term of the GP is \(3\).

Let the three numbers in AP be \(a - d,~ a, ~a + d ~ ( d >0)\).

Given,

The largest number is \(7\) times the smallest.

So, \(a+d=7(a-d)\)

\(\Rightarrow a+d=7 a-7 d\)

\(\Rightarrow 8 d =6 a\)

\(\Rightarrow d=\frac{3 a}{ 4}\)

The product of three numbers in AP \(=224\)

\(\therefore (a-d)\times a \times (a+d)=224\)

\(\Rightarrow a \left( a ^{2}- d ^{2}\right)=224 \)...(1)

Putting the value of \(d\) in equation (1), we get

\( a\left(a^{2}- \left(\frac{3 a}{4}\right)^{2}\right)=224\)

\(\Rightarrow a\left(a^{2}-\frac{9 a^{2}}{16}\right)=224\)

\(\Rightarrow a\left(\frac{16 a^{2}-9 a^{2}}{16}\right)=224\)

\(\Rightarrow a \times 7 a ^{2}=224 \times 16\)

\(\Rightarrow 7 a ^{3}=224 \times 16\)

\(\Rightarrow a ^{3}=\frac{224 \times 16}{7}\)

\(\Rightarrow a^{3}=32 \times 16\)

\(\Rightarrow a^{3}=512\)

\(\Rightarrow a^{3}=8^{3}\)

As bases are same,

\(\therefore a =8\)

\(d =\frac{3 a }{ 4}=\frac{3 \times 8}{ 4}=6\)

\(a - d =8-6=2\)

\(a + d =8+6=14\)

So, the three numbers are \(2,8,14\) and in these numbers \(14\) is the largest number.

Term of an AP are \(403,397,391 \ldots \)

So, First term \((a) =403\)

Common difference \((d)=397-403=-6\)

Let \(n ^{\text {th }}\) term is the first negative term of the given AP.

As we know,

\(n^{\text {th}}\) term of AP is given by,

\(a_ n=a+(n-1) \times d\)

\(\therefore a_n =403+(n-1) \times-6 \)

\(\Rightarrow a_n=403-6 n+6 \)

\(\Rightarrow a_n=409-6 n\)

Now, we have to find the suitable value of \(n\), so that the value of \((409-6 n)\) is negative.

Checking options:

Taking \(n=68\)

\(\therefore a_{n}=409-6 \times 68\)

\(\Rightarrow a_n=409-408\)

\(\Rightarrow a_n=1\) (which is not negative)

Now, putting \(n=69\), we get

\(\therefore a_{n}=409-6 \times 69\)

\(\Rightarrow a_n=409-414\)

\(\Rightarrow a_n=-5\)

This is the first negative term of the series.

So, \(69^{\text {th}}\) term is the first negative term of given series.

Given,

The geometric mean of two non-negative observations \(=10\)

The harmonic mean of two non-negative observations \(=8\)

As we know,

Relation between Arithmetic mean (AM),), Geometric mean (GM) and Harmonic mean (HM) is given by:

\((GM)^2= (AM) \times (HM)\)

\(\therefore (10)^2=AM \times 8\)

\(\Rightarrow 100= AM \times 8\)

\(\Rightarrow \frac{100}{8}=AM\)

\(\therefore AM= 12.5\)

So, the arithmetic mean of the observations is \(12.5\).