Sum of first n natural numbers

The marks in the first 15 steps are 1 through 15, so their sum is \(\frac{15\times16}{2}=120\). The next 10 steps contribute zero marks, so they do not change the total. Option 130 would require 10 additional marks, but these steps add zero, not 10. Exam tip: use \(\frac{n(n+1)}{2}\) for the sum of natural numbers from 1 to n.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{52}=\frac{52\times53}{2}=26\times53=1378\). Hence, option B is correct. Option A is a close distractor because it equals the sum up to \(51\): \(\frac{51\times52}{2}=1326\). Exam tip: for the sum of consecutive natural numbers from 1 to \(n\), use \(n(n+1)/2\).

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. Thus, \(S_{33}=\frac{33\times34}{2}=561\) and \(S_{22}=\frac{22\times23}{2}=253\). Therefore, \(S_{33}-S_{22}=561-253=308\). It is also the sum of the integers from \(23\) to \(33\); an option such as \(298\) can result from an incorrect term count or sum. Exam tip: interpret \(S_m-S_n\) as the sum from \((n+1)\) to \(m\).

The direct answer is option C, 45. The sum of the first n natural numbers is S_n=n(n+1)/2. Set this equal to 1035: n(n+1)/2=1035, so n(n+1)=2070. Testing option C gives 45·46=2070, hence S_45=45·46/2=45·23=1035. Therefore n=45. Option A gives S_43=43·44/2=946, not 1035. Option B gives S_44=44·45/2=990, also not 1035. Option C gives exactly the required sum. Option D gives S_46=46·47/2=1081, which is too large. Since the sum increases as n increases, once 44 gives less and 46 gives more, 45 is confirmed. A useful check is that the sum of the first 45 numbers is 1+45, 2+44, and so on: 22 pairs of 46 plus the middle 23, giving 1012+23=1035. Memory cue: triangular-number formula is n(n+1)/2.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. Adding 20 to \(1+2+3+\cdots+19\) gives \(1+2+3+\cdots+19+20\), which is \(S_{20}\). \(S_{19}\) includes terms only up to 19, so it is not correct. Exam tip: use the last included natural number to identify the subscript of the sum.

Total saving is (S_{31}=\frac{31\times32}{2}=496) rupees. A daily increase of (1) forms the sum of natural numbers.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Thus, \(S_{18}=\frac{18\times19}{2}=171\). Three times this sum is \(3\times171=513\), so 513 is correct. A nearby value such as 503 results from not applying the sum formula correctly. Exam tip: first find the sum using \(\frac{n(n+1)}{2}\), then apply the required multiplication.

The sum of the first 26 natural numbers is
\(S_{26}=\frac{26\times27}{2}=351\). Therefore, subtracting 26 gives
\(351-26=325\). This is also the sum of the first 25 natural numbers. For example, 315 is not the correct sum of the first 25 natural numbers. Exam tip: remember that the sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\).

These are 10 consecutive natural numbers from 16 to 25. Their average is \(\frac{16+25}{2}=20.5\), so the sum is \(10\times20.5=205\). A value such as 195 can result from counting the number of terms or the average incorrectly. Exam tip: for consecutive terms, first count them using \((25-16+1)\).

The sum of the first n natural numbers is
\(S_n=1+2+\cdots+n\), whereas
\(S_{n-1}=1+2+\cdots+(n-1)\). Therefore,
\(S_n-S_{n-1}=n\). Since the given difference is 37,
\(n=37\). For n = 36 or 38, the difference would be 36 or 38 respectively. Exam tip: the difference between two consecutive partial sums is the newly added term.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{55}=\frac{55\times56}{2}=55\times28=1540\). Hence, 1540 is the correct option. The value \(1485\) is the sum up to 54, since \(\frac{54\times55}{2}=1485\). Exam tip: divide the even factor by 2 first to calculate quickly.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. Therefore, \(S_{25}=S_{23}+24+25=276+49=325\). Hence, the correct option is \(325\). A value such as \(300\) can result from the mistake of adding only one new term. Exam tip: When moving from \(S_n\) to \(S_{n+k}\), add every intervening new term.

If \(S_n\) denotes the sum of the first \(n\) natural numbers, then adding the next two numbers to \(S_{29}\) gives \(S_{29}+30+31=S_{31}\). \(S_{30}\) includes only up to 30, so it is not correct. Exam tip: when consecutive next terms are added to a partial sum, increase the subscript accordingly.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Thus, \(S_{20}=\frac{20\times21}{2}=210\) and \(S_{10}=\frac{10\times11}{2}=55\). Hence, \(S_{20}:S_{10}=210:55=42:11\). The ratio \(21:5\) would result from incorrectly taking \(S_{10}\) as 50. Exam tip: simplify a ratio at the end by dividing both terms by their greatest common factor.

Total beads are (S_{19}=\frac{19\times20}{2}=190). Triangular rows often give the sum of natural numbers.

The difference of two consecutive sums is the new term, so (S_{41}-S_{40}=41). You can answer without finding the full sums.

The sum of the first r natural numbers is \\(\frac{r(r+1)}{2}\\). Thus, \\(\frac{r(r+1)}{2}=153\\), so \\(r(r+1)=306=17\times18\\) and \\(r=17\\). Therefore, \\(r+2=17+2=19\\). Option 18 is only the value of \\(r+1\\), not \\(r+2\\). Exam tip: when a sum is given, first use \\(\frac{n(n+1)}{2}\\) to find n.

The sum of the first \(n\) natural numbers is \(S_n=\frac{n(n+1)}{2}\). Therefore, \(S_{46}=\frac{46\times47}{2}=23\times47=1081\). Hence, 1081 is correct. An option such as 1071 can result from an arithmetic error in multiplication or addition. Exam tip: for the sum of consecutive natural numbers, directly use \(\frac{n(n+1)}{2}\).

Use the formula S_n = n(n + 1)/2 for the sum of the first n natural numbers. Thus S₁₂ = 12×13/2 = 78, S₁₃ = 13×14/2 = 91, and S₁₄ = 14×15/2 = 105. Their total is 78 + 91 + 105 = 274, so option A is correct. The original listed options and answer were inconsistent; they have been corrected here.

The sum of the first n natural numbers is \(S_n=\frac{n(n+1)}{2}\). From \(\frac{a(a+1)}{2}=351\), we get \(a=26\), since \(\frac{26\times27}{2}=351\). Similarly, \(\frac{b(b+1)}{2}=465\) gives \(b=30\), since \(\frac{30\times31}{2}=465\). Therefore, \(b-a=30-26=4\). Option 3 is incorrect because it is one less than the actual difference between the two indices. Exam tip: equate each given sum to \(\frac{n(n+1)}{2}\) to find its index.

The sum of the first \(n\) natural numbers is \(\frac{n(n+1)}{2}\). Substituting \(n=70\), we get \(\frac{70\times71}{2}=35\times71=2485\). Therefore, 2485 is correct. A value such as 2475 results from an arithmetic error in multiplication or addition. Exam tip: cancel 2 with an even factor in \(n(n+1)\) before multiplying.

Direct answer: n = 24, so option B is correct. For the first n natural numbers, Sₙ = n(n + 1)/2. Put Sₙ = 300: n(n + 1)/2 = 300, which gives n(n + 1) = 600. The value n = 24 works because 24 × 25 = 600. Therefore S₂₄ = 24 × 25/2 = 12 × 25 = 300. If an algebraic solution is preferred, expand to n² + n − 600 = 0. This factors as (n + 25)(n − 24) = 0, giving n = −25 or n = 24. Since n counts a number of terms, it must be a positive natural number, so −25 is rejected. The other choices fail: S₂₃ = 276, S₂₅ = 325, and S₂₆ = 351. Memory cue: after multiplying the given sum by 2, look for two consecutive factors whose product is that number.

(S_{37}=703) and (S_{38}=741), so the sum is (1444). Notice the one-term difference in nearby sums.

The governing concept is the sum of the first n natural numbers and cancellation of common terms. Let Sₙ = 1 + 2 + ... + n. In S₆₄ − S₆₀, the terms 1 through 60 occur in both sums and cancel, leaving 61 + 62 + 63 + 64. Pairing the remaining terms gives (61 + 64) + (62 + 63) = 125 + 125 = 250. The standard formula confirms this: S₆₄ = 64×65/2 = 2080 and S₆₀ = 60×61/2 = 1830, so their difference is 2080 − 1830 = 250. Hence option C is correct. The other options can result from omitting a remaining term or making an arithmetic error.

Here, \(S_n\) denotes the sum of the first \(n\) natural numbers. To move from \(S_{52}\) to \(S_{54}\), add the next two numbers, 53 and 54: \(S_{54}=S_{52}+53+54=1378+107=1485\). Therefore, 1485 is correct. Option 1495 would require an increase of 117, whereas the required increase is \(53+54=107\). Exam tip: In questions involving nearby values of \(S_n\), adding the new terms is often quicker than applying the full formula.