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The different program errors are as follows;

1. Syntax error:

An error occurs when there is a violation of the grammatical rules of a programming language’s instructions. It happens at the time of compilation. Such errors need to be rectified before proceeding further.

2. Semantic errors:

An error, which occurs due to the incorrect logic in a solution is called semantic error. It also occurs due to the wrong use of grammar in the program.

3. Runtime Errors:

occur at run-time. Such an error causes a program to end abruptly or even cause system shut-down. Such errors are hard to detect and are known as ‘Bugs’.

4. Logical Error:

It may happen that a program contains no syntax or run-time errors but still, it doesn’t produce the correct output. It is because the developer has not understood the problem statement properly. These errors are hard to detect as well. It may need the algorithm to be modified in the design phase and changing sources code.

Top-Down Approach:

It is based on a concept called divide and conquer. A given problem is solved by breaking it down into smaller manageable parts called modules. Hence it is also called as step wise refinement. The subprograms are further divided into still smaller sub problems. Finally, the sub problems are solved individually, and all these give the solution to the overall problem.

Properties of Top-Down Analysis:

1. Understandability:

The individual modules are organized to execute in a particular sequence.

2. This helps to understand the program behaviour more easily.

3. Clear Identification of tasks.

4. Easy program maintenance.

5. Removes duplication or repetition of coding in a problem.

6. Enhances the feature of code reusability.

1. c. Coorg is situated between Mysore and the coastal town of Mangalore.

2. d. Spices and coffee grow in Coorg in plenty.

3. b. The best period to visit Coorg is between September and March.

4. a. The author says that the Coorg is inhabited by a proud race of martial men and beautiful women.

5. a. Coorg is also known as Kodagu.

We have,

\(0.0.3\overline{56}\) = 0.3 + 0.056 + 0.00056 + 0.0000056 + ⋯ ∞

⇒ \(0.3\overline{56}\) = 0.3 + \(\big[\frac{56}{10^3} + \frac{56}{10^5} + \frac{56}{10^7} + ⋯ ∞\big]\)

⇒ \(0.3\overline{56}\) = \(\frac{3}{10}\) + \(\frac{\frac{56}{10^3}}{1-\frac{1}{10^2}}\) = \(\frac{3}{10}\) + \(\frac{56}{990}\) = \(\frac{353}{990}\)

Let the three numbers in G.P. be a, ar, and ar².

From the given condition,

a + ar + ar² = 56

⇒ a (1 + r + r²) = 56 ……………………….… (1)

a – 1, ar – 7, ar² – 21 forms an A.P.

∴ (ar – 7) – (a – 1) = (ar² – 21) – (ar – 7)

⇒ ar – a – 6 = ar² – ar – 14

⇒ar² – 2ar + a = 8

⇒ar² – ar – ar + a = 8

⇒a(r² + 1 – 2r) = 8

⇒ a (r – 1)² = 8 ……………………………… (2)

From (1) and (2), we get

⇒7(r² – 2r + 1) = 1 + r + r²

⇒7r² – 14 r + 7 – 1 – r – r² = 0

⇒ 6r² – 15r + 6 = 0

⇒ 6r² – 12r – 3r + 6 = 0

⇒ 6r (r – 2) – 3 (r – 2) = 0

⇒ (6r – 3) (r – 2) = 0

When r = 2, a = 8

When

Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.

When, r=1/2, the three numbers in G.P. are 32, 16, and 8.

Thus, in either case, the three required numbers are 8, 16, and 32.

(i) nth term of the given series

Tn = 1² + 2² + 3² + ..... + n² = ∑n² = \(\frac{n(n+1)(2n+1)}{6}\) = \(\frac{2n^3+3n^2+n}{6}\) = \(\frac{n^3}{3}\) + \(\frac{n^2}{2}\) + \(\frac{n}{6}\)

∴ S_n = \(\frac{1}{3}\). ∑n³ + \(\frac{1}{2}\).∑n² + \(\frac{1}{6}\). ∑n = \(\frac{1}{3}\).\(\frac{n^2(n+1)^2}{4}\) + \(\frac{1}{2}\).\(\frac{n(n+1)(2n+1)}{6}\) + \(\frac{1}{6}\).\(\frac{n(n+1)}{4}\)

= \(\frac{n(n+1)}{12}\) {n(n + 1) + (2n + 1) + 1} 

= \(\frac{n(n+1)(n^2+3n+2)}{12}\) = \(\frac{n(n+1)(n+1)(n+2)}{12}\) 

= \(\frac{n^2(n+1)^2(n+2)}{12}\)

(ii) In the given series each term is the product of two factors. The factors 3, 4, 5, ..... are in A.P. having 3 as the first term and 1 as the common difference, therefore the nth term of this A.P. = 3 + (n – 1)1 = 2 + n. Also, the factors 5, 7, 9, .... are in A.P. having 5 as the first term and 2 as the common difference, therefore the nth term of this A.P. = 5 + (n – 1)² = 2n + 3

∴ nth term of the series = (2 + n) (2n + 3) 

⇒ T_n = 2n² + 7n + 6

∴ S_n = 2 . ∑n² + 7∑n + 6n = \(\frac{2n(n+1)(2n+1)}{6}\) + \(\frac{7n(n+1)}{2}\) + 6n

= \(\frac{2n(n+1)(2n+1)+21n(n+1)+36n}{6}\) = \(\frac{4n^3+27n^2+59n}{6}\)

Let 

S = 1³ + 2³ + 3³ + 4³ + ..... + n³ 

Then, S = ∑n³ = \(\bigg\{\frac{n(n+1)}{2}\bigg\}^2\) = (∑n)²

(i) Varies inversely.

Since the time taken to construct the wall varies with the number of men hired. If the number of men hired is increased then time required for constructing wall shall decrease and vice-versa.

(ii) Varies directly.

Since the cost of bus journey will vary with distance. If the distance of journey will increase then cost of ticket will increase and vice-versa.

(iii) Varies directly.

Since the quantity of petrol required varies with the distance. If the distance of journey will increase then petrol required will also increase and vice-versa.

Correct option is (D) 2

The given sequence is

\(\frac{1}{2}\) + \(\frac{3}{4}\) + \(\frac{7}{8}\) + \(\frac{15}{16}\) + ⋯ + n terms

We can write each individual as,

\(\frac{1}{2}\) = 1 − \(\frac{1}{2}\)

\(\frac{3}{4}\) = 1 − \(\frac{1}{4}\)

\(\frac{7}{8}\) = 1 − \(\frac{1}{8}\) and … till n terms

Now writing each term in its new form, we get,

\(\big(1-\frac{1}{2}\big)+\big(1-\frac{1}{4}\big)+\big(1-\frac{1}{8}\big)+...+\big(1-\frac{1}{2_n}\big)\)

= (1 + 1 + 1 … n terms) − \(\big(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\,...+\frac{1}{2_n}\big)\)

= n − \(\big(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\,...+\frac{1}{2_n}\big)\) … (i)

Now, \(\big(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\,...+\frac{1}{2_n}\big)\) is a G.P with common ratio = \(\frac{1}{2}\) so,

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\,...+\frac{1}{2_n}\) = \(\frac{\frac{1}{2} \big(1 − (\frac{1}{2})^n\big)}{1 − \frac{1}{2}}\)

= \(\bigg[1 −\big (\frac{1}{2}\big)^n \bigg]\)

Putting this value in eq.(i), we get

n − \(\big(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\,...+\frac{1}{2_n}\big)\)

= n − \(\bigg[1 −\big (\frac{1}{2}\big)^n \bigg]\) = n − 1 + \(\big(\frac{1}{2}\big)^n\)

\(\displaystyle\sum_{i=1}^{10} (2r-1)^2\) = 1² + 3² + ⋯ + 17² + 19²

To find the sum of above 10 terms, we find the sum of n terms in general.

∴ Let T_n be n^th term of the series.

1² + 3² + 5² + 7² + ..... to n terms

Now, T_n = [1 + (n − 1) × 2]² = 4n² − 4n + 1

∴ 1² + 3² + 5² + ⋯ to n terms

= \(\displaystyle\sum_{k=1}^{n} T_k\)

= \(\displaystyle\sum_{k=1}^{n} (4k^2-4k'1)\)

= \(4\bigg(\displaystyle\sum_{k=1}^{n} k^2\bigg)-4\bigg(\displaystyle\sum_{k=1}^{n} k\bigg)+\) \(\bigg(\displaystyle\sum_{k=1}^{n} 1\bigg)\)

= 4 \(\bigg[\frac{n(n + 1)(2n + 1)}{6}\bigg]\) − 4 \(\bigg[\frac{n(n + 1)}{6}\bigg]+n\)

= \(\frac{n}{3}\)[2(n + 1)(2n + 1) − 6(n + 1) + 3]

= \(\frac{n}{3}\)[4n² + 6n + 2 − 6n − 6 + 3]

= \(\frac{n}{3}\)(4n² − 1)

Now put n = 10 in the above result, we get

1² + 3² + 5² + ⋯ + 17² + 19² = \(\frac{10}{3}\) (4.10² − 1)

= \(\frac{10}{3}\)(399)

= 1330

Let T_n be the n^th term of the given series,

Then

T_n = (n^th term of 1,5,9 …) × (n^th term 3,7,11 …) × (n^th term of 4,8,12 …)

= [1 + (n − 1)4] × [3 + (n − 1)4] × [4 + (n − 1)4]

= (4n − 3) × (4n − 1) × (4n)

= (16n² − 16n + 3)4n

= 64n³ − 64n² + 12n

Let S_n denote the sum to n term of the given series. Then,

S_n = \(\displaystyle\sum_{k=1}^{n} T_k = \displaystyle\sum_{k=1}^{n}(64n^3-64n^2+12n)\)

= \(64\displaystyle\sum_{k=1}^{n} n^3 -64\displaystyle\sum_{k=1}^{n}n^2+12\displaystyle\sum_{k=1}^{n}n\)

= 64 \(\big[\frac{n(n + 2)}{2} \big]^2\) − 64 \(\big[\frac{n(n + 1)(2n+1)}{6} \big]\) + 12 \(\big[\frac{n(n + 1)}{2} \big]\)

= \(\frac{64\{n(n + 1)\}^2}{4}\) − \(\frac{64\{n(n + 1)(2n+1)\}}{6}\) + 6n(n + 1)

= 16{n(n + 1)}² − \(\frac{32\{n(n + 1)(2n+1)\}}{3}\) + 6n(n + 1)

= \(\frac{n(n + 1)}{3} \)[48{n(n + 1)} − 32(2n + 1) + 18]

= \(\frac{n(n + 1)}{3} \)[48(n² + n) − (64n + 32) + 18]

= \(\frac{n(n + 1)}{3} \)[48n² + 48n − 64n − 32 + 18]

= \(\frac{n(n + 1)}{3} \) [48n² − 16n − 14]

Let, S = 7 + 77 + 777 + …77 … 7 (upto n terms)

S = 7[1+ 11 + 111 + …11 … 1] (upto n terms)

s = \(\frac{7}{9}\)[9 + 99 + 999 + ⋯ 99 … 9]

s = \(\frac{7}{9}\)[10 − 1 + 10² − 1 + 10³ − 1 + ⋯ 10ⁿ − 1]

s = \(\frac{7}{9}\)[10¹ + 10² + ⋯ 10ⁿ − n]

s = \(\frac{7}{9}\)\(\big[\frac{10(10^n − 1)}{10 − 1} − n\big]\)

s = \(\frac{7}{9}\)\(\big[\frac{10}{9}(10^n-1) − n\big]\)

Let

S_n = 1 + (1 + x) + (1 + x + x²) + (1 + x + x² + x³ ) + ⋯

to n terms

Hence, (1 + x) S_n = (1 + x) + (1 − x)(1 + x)

+ (1 − x) + (1 + x + x²)

+ (1 − x) + (1 + x + x² + x³) + ⋯

to n terms

or (1 − x) S_n = (1 − x) + (1 − x²) + (1 − x³)

to n terms + (1 − x⁴) + ⋯

= n − (x + x² + x³ + x⁴ + ⋯ ) to n terms

= n − \(\frac{x(1 − x^n)}{(1 − x)}\)

Hence, S_n = \(\frac{n}{1-x}\) − \(\frac{x(1 − x^n)}{(1 − x)^2}\)

We have,

1 + 3 + 3² + ⋯ + 3ⁿ⁻¹ > 1000

⇒ 3⁰ + 3¹ + 3¹ + ⋯ 3ⁿ⁻¹ > 1000

⇒ 3⁰\(\big(\frac{3^n − 1}{3 − 1}\big)\) > 1000

⇒ 3ⁿ − 1 > 2000

⇒ 3ⁿ > 2001

Least value of n, which satisfies this inequality is n = 7 (∵ 3⁷ = 2187)

Hence, least value of n = 7.

(3) 0 ≤ r < b 

100 + 98 + 96 + 94 + … 30 steps

Here 

a = 100

d = -2 

n = 30 

∴ S_n = (n/2) (2a + (n – 1)d) 

S₃₀ = (30/2) (2 x 100 + 29 x -2) 

= 15(200 – 58) 

= 15 x 142 

= 2130 

t₃₀ = a + (n – 1)d 

= 100 + 29 x -2 

= 100 – 58 

= 42

(i) No. of bricks required for the top step are 42. 

(ii) No. of bricks required to build the stair case are 2130.

Answer is (b) (e^{sin √x}.cos √x)/2√x

Answer is (c) -1/2√x sin √x

x-α/0 = y-β/0 = z-γ/1

The Rajasthan Central India Assembly established in 1927 had main purpose to introducing people to the Congress activities.

The assembly was organised by Seth Jamanlal Bajaj; Vijay Singh Pathik and other representatives jointly.

Maharana Fateh Singh was the Maharana of Mewar during Delhi Court in 1911.

(c) Gurumukh N. Singh

Mewar Maharana signed the merger letter on 11th April 1948.

On 18 March 1948, the merger of Alwar, Bharatpur, Dholpur and Karoli provinces formed Matsya Confederation.

During the British era, Ajmer – Merwada was a centrally ruled state. All India Nationalist State Public Council’s Rajputana Regional Assembly always demanded that the locality of Ajmer Merwada be includedin Greater Rajasthan along with other regions.

After the 1952 AD General elections in Ajmer Merwada, the Congress Ministry was formed in the leadership of Shri Hari Bhau Upadhyaya. The Congress leadership was not in favour of , merger of Ajmer in Rajasthan and now after the formation of ministry in Ajmer Merwada, the Congress leadership gave the logic that from the perspective of administration, it remained a small state.

This report was handed over to State Reorganisation Committee. Thereafter on 1 November, 1956, Sirohi’s Mount Abu area and alongwith this, Ajmer Merwada was also included in Rajasthan.

Greater Rajasthan’s Prime Minister was Hira Lal Shastri who took charge on 4 : April 1950.

(a) Gosul Bhai Bhatt

Hari Narayan Sharma founded untouchable organisation, Valmiki organisation, publicised Khadi and worked for religious solidarity and people’s awareness.

During British era, Ajmer – Marwada was a centrally-administered state.

Correct Answer is: (b) Jaipur

The gathering was held in the presidentship of Jaynarayan Vyas.

This convention was presided over by Jawahar Lal Nehru.

The foremost reason of public awareness in Rajasthan was Swami Dayanand Saraswati and his influence. He was the first social reformer to blow the trumpet for selfmade products and self rule. The second reason was newspapers and literature. Rajputana gazette, ‘Rajasthan Kesari’ and ‘Naveen Rajasthan’ were the main newspapers. The role of middle class was another reason. In this, educationist, advocate and journalist Jai Narayan Vyas, Master Bhola Nath were prominent.

The armies of almost all the states of Rajasthan participated in the 1st world war. The armymen shared their own experiences and introduced the people of Rajasthan to nationalistic ideas. The external environment’s influence was a major reason of public awareness. The national leaders and their programmes also influenced people.

(d) Matsya confederation

Bhoop Singh.

The revolution of 1857 was primarily an outpouring of public resentment against the British and the oppressive feudals armies. Mainly 6 army cantonments were famous. These were Naseerabad, Neemuch, Devli, Kota, Erinpura, Kherwada. The revolution of Neemuch via Devli reached Kota. The revolutionaries compelled Kota regiment’s 60 people to come alongwith them from Devli cantonment but on the way the armymen were successful in running away.

On 15th October Kota’s Maharao Bhim Singh’s two platoons revolted. Resident Burton was beheaded and Maharao was made captive and the revolutionaries gained control over whole of Kota. The role of Maharao’s advocate and intellectual Jai Dayal Bhatnagar in the organised public outcry was an important contribution.

(a) Matsya confederation

Doongi and Jawaharji both looted British cantonments and government treasury and helped the poor. That is why they were popular among people.

In Kota, the British political agent Major Burton was killed.

Salumbar’s Rawal Kesri Singh helped Tatya Tope.

(d) Kesari Singh Barhat.

Before the 1857 revolt, Jodhpur’s Man Singh demonstrated emotions against the British.

Auwa was the main centre of revolt against the Britishers. The ruler of this place Kushal Singh gave significant help to revolutionaries.

Jaipur’s ruler Ram Singh helped Britishers in this revolt.

(b) the soldiers of Naseerabad were asked to use bullets coated with cow fat

Veer Satsai was written by Suryamal Mishra.

(a) Swaroop Singh

(c) Vijay Singh Pathik