Sum of first n natural numbers

Correct answer: C. 41

Given \(\frac{n(n+1)}{2}=861\), we get \(n(n+1)=1722\). Since \(41\times42=1722\), \(n=41\). For the close distractor \(40\), the sum is \(\frac{40\times41}{2}=820\), not 861. Exam tip: look for two consecutive numbers whose product equals twice the given sum.

Correct answer: B. 950

To find the sum from 26 to 50, subtract the sum of natural numbers up to 25 from the sum up to 50. Thus, \\(S_{50}-S_{25}=\frac{50\times51}{2}-\frac{25\times26}{2}=1275-325=950\\). Therefore, the correct answer is 950. An option such as 975 can result from using an incorrect number of terms or lower limit. Exam tip: For a range sum, always subtract the sum up to one less than the lower limit.

Correct answer: D. 64

Here, \(S_n-S_{n-4}\) represents the sum of the four natural numbers from the \((n-3)\)th term to the \(n\)th term: \(S_n-S_{n-4}=(n-3)+(n-2)+(n-1)+n=4n-6\). Hence, \(4n-6=250\), so \(4n=256\) and \(n=64\). Therefore, 64 is the correct option. For instance, if \(n=63\), the sum is \(60+61+62+63=246\), not 250. Exam tip: \(S_n-S_{n-k}\) always gives the sum of the last \(k\) terms.

Correct answer: A. \(1010\)

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{60}=\frac{60\times61}{2}=1830\) and \(S_{40}=\frac{40\times41}{2}=820\). Therefore, the difference is \(1830-820=1010\). It is also the sum of the numbers from \(41\) to \(60\); \(1000\) does not give the correct difference. Exam tip: the difference of two such sums can be found by adding the remaining consecutive terms directly.

Correct answer: C. 1431

Direct answer: option C, 1431. Since Sₙ is the sum of the first n natural numbers, Sₙ = n(n + 1)/2. The given equation gives n(n + 1)/2 = 1275, so n(n + 1) = 2550. The consecutive numbers 50 and 51 have product 2550, hence n = 50. The required sum is therefore Sₙ₊₃ = S₅₃. Applying the formula again, S₅₃ = 53 × 54/2 = 53 × 27 = 1431. Option A, 1378, does not equal the sum through 53. Option B, 1406, is S₅₂ and reflects increasing the index by only two. Option C is correct because the index must increase from 50 to 53. Option D, 1485, comes from using an incorrect index or arithmetic. Memory cue: first solve for the original index, then apply the requested shift; do not shift the sum value itself.

Correct answer: B. जब n को 4 से विभाजित करने पर शेषफल 1 या 2 हो

The sum is \(S_n=\frac{n(n+1)}{2}\). It is odd only when n leaves remainder 1 or 2 on division by 4. Merely knowing that n is odd or even is not enough. Exam tip: check \(n\bmod 4\).

Correct answer: A. 1575

There are \(75-51+1=25\) terms from 51 to 75, inclusive. This is an arithmetic sequence with first term 51 and last term 75. Hence, the sum is \(\frac{25}{2}(51+75)=\frac{25}{2}\times126=1575\). Therefore, option A is correct. An answer such as 1600 can result from an error in counting the terms or finding the average. Exam tip: always add \(+1\) when counting terms from one inclusive endpoint to another.

Correct answer: B. 1653

For Sₙ = 1 + 2 + 3 + ... + n, subtracting S₍ₙ₋₁₎ = 1 + 2 + ... + (n − 1) leaves only the final term n. Hence Sₙ − S₍ₙ₋₁₎ = n. The given difference is 57, so n = 57. Now calculate the required sum: S₅₇ = 57 × 58/2 = 57 × 29 = 1653. Therefore option B is correct. Option A is close but reflects an arithmetic error in the product or division. Options C and D correspond to using a larger index or adding an unrelated term. The important concept is that the difference of two consecutive partial sums identifies the newly included natural number.

Correct answer: B. 65:11

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{25}=\frac{25\times26}{2}=325\) and \(S_{10}=\frac{10\times11}{2}=55\). Therefore, the ratio is \(325:55=65:11\). Option 55:13 is incorrect because it is not obtained by simplifying 325:55 correctly. Exam tip: always reduce a ratio by dividing both terms by their greatest common factor.

Correct answer: B. 340

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. From \(\frac{n(n+1)}{2}=2485\), we get \(n=70\), since \(\frac{70\times71}{2}=2485\). Therefore, \(S_n-S_{n-5}=S_{70}-S_{65}\), which is the sum of the last five numbers: \(66,67,68,69,70\). Hence, \(66+67+68+69+70=340\). Choosing 350 would miss the correct first term, 66, among these five terms. Exam tip: \(S_n-S_{n-r}\) always represents the sum of the last \(r\) terms.

Correct answer: A. 1965

The sum of the first 80 natural numbers is \(\frac{80\times81}{2}=3240\), and the sum of the first 50 natural numbers is \(\frac{50\times51}{2}=1275\). Therefore, the required difference is \(3240-1275=1965\). Equivalently, it is the sum of the integers from 51 to 80. Exam tip: use \(n(n+1)/2\) carefully for both sums to avoid a close error such as 1975.

Correct answer: A. 225

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{n(n+1)}{2}=595\), we get \(n=34\). Therefore, \(S_{n+6}-S_n=S_{40}-S_{34}=35+36+37+38+39+40=225\). Although 230 may seem close, it does not equal the sum of the next six terms. In exams, first find n and then add only the newly included terms.

Correct answer: B. 465

For \(n=15\), the index \(2n\) becomes 30. Hence, \(S_{2n}=S_{30}=\frac{30(30+1)}{2}=\frac{30\times31}{2}=465\). Option 435 equals \(S_{29}\), so it is not correct. Exam tip: evaluate the index \(2n\) before substituting into the formula.

Correct answer: C. \(S_{2n}=2S_n+n^2\)

\(S_{2n}=\frac{2n(2n+1)}{2}=n(2n+1)\), whereas \(2S_n=n(n+1)\). Their difference is \(n^2\), so \(S_{2n}=2S_n+n^2\). Option D ignores this extra \(n^2\) term. Exam tip: substitute \(S_n=\frac{n(n+1)}{2}\) before comparing such expressions.

Correct answer: C. (210)

Adding the two equations gives (2S_n=420), so (S_n=210). When two equations are given, addition is quick.

Correct answer: C. It is half the product of two consecutive integers.

The sum of the first n natural numbers is \(1+2+\cdots+n=\frac{n(n+1)}{2}\), half the product of consecutive integers n and \(n+1\). It is not always even: for n=2, the sum is 3. Exam tip: check the \(n+1\) factor carefully.

Correct answer: A. 222

The sum of the first 36 natural numbers is \(S=\frac{36\times 37}{2}=666\). Therefore, \(\frac{1}{3}\) of this sum is \(666\div 3=222\). Although 224 is close, it is not obtained by dividing 666 by 3. Exam tip: first use \(\frac{n(n+1)}{2}\) to find the sum, then apply the given fraction.

Correct answer: B. (56)

(S_{56}-S_{30}=1596-465=1131), so (1110) does not match the listed options. Always check options carefully.

Correct answer: B. 2210

There are 20 numbers from 101 to 120, inclusive. Therefore, using the sum of an arithmetic progression,
\(S=\frac{20}{2}(101+120)=10\times221=2210\). Hence, option B is correct. The value 2190 is the sum from 100 to 119, so it does not match the given range. Exam tip: Count terms in an inclusive range using \(\text{last term}-\text{first term}+1\).

Correct answer: A. 4095

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Since \(5050=\frac{100\times101}{2}\), we get \(n=100\). Therefore, \(S_{n-10}=S_{90}=\frac{90\times91}{2}=4095\). Option 4186 is not correct because it is the sum of the first 91 natural numbers. Exam tip: Recognise 5050 as \(S_{100}\), then reduce the subscript by 10.

Correct answer: A. 69:31

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{45}=\frac{45\times46}{2}=1035\) and \(S_{30}=\frac{30\times31}{2}=465\). Therefore, \(S_{45}:S_{30}=1035:465=69:31\). The option \(45:30\) is only the ratio of the numbers of terms, not the ratio of their sums. Exam tip: For ratios of such sums, apply \(\frac{n(n+1)}{2}\) first and then cancel common factors.

Correct answer: B. 1326

Here, \(S_n\) is the sum of the first \(n\) natural numbers. In \(S_n-S_{n-2}\), the first \(n-2\) terms cancel, leaving \((n-1)+n=2n-1\). Thus, \(2n-1=101\) gives \(n=51\). Therefore, \(S_{51}=\frac{51\times52}{2}=1326\). Option 1275 is \(S_{50}\), so it is close but not correct. Exam tip: In differences of partial sums, cancel the common initial terms and write the remaining terms.

Correct answer: B. 930

There are \(45-15+1=31\) terms from 15 to 45. The sum of the first and last terms is \(15+45=60\), so the sum is \(\frac{31\times 60}{2}=930\). Although 960 is close, it does not follow from the correct number of terms and average of this sequence. Exam tip: When both endpoints are included, always use \(+1\) while counting the terms.

Correct answer: B. \(2n+1\)

\(S_{n+1}=1+2+\cdots+n+(n+1)\) and \(S_{n-1}=1+2+\cdots+(n-1)\). On subtracting, the common terms cancel, leaving \(n+(n+1)=2n+1\). Hence, the correct answer is \(2n+1\). \(n+1\) is only the newly added last term; the term \(n\) must also be included. Exam tip: Expand the two sums and identify the terms that cancel.

Correct answer: A. 1452

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{72}=\frac{72\times73}{2}=2628\) and \(S_{48}=\frac{48\times49}{2}=1176\). Hence, the difference is \(2628-1176=1452\). It is also the sum of the numbers from 49 to 72; an option such as 1462 results from an arithmetic error. Exam tip: for such questions, write \(S_{72}-S_{48}\) directly.