Arithmetic Progressions: questions with answers

Solving $20 - (n-1)(3) = -101$ gives $(n-1)(3) = 121$, so $n - 1 = 40$ with $1$ left over -- $n$ comes out between $41$ and $42$ and is not a whole number at all. Since $n$ counts terms, that IS the answer: $-101$ is simply not a term of this AP. Rounding to the nearest position converts a clean no into a confident wrong yes, and quoting $-101$ answers with the value the question gave rather than with a position.

The list itself FALLS by $3$, so reading it backwards RISES by $3$: the $11$th from the end is $-70 + 10 \times 3 = -40$. Two directions pull opposite ways here. Stepping from the last term in the list's own direction gives $-100$, and counting from the front gives $-10$.

The step count is $\frac{ 127 - 7 }{ 5 } = 24$, which is how many times $5$ has been added to get from the first term to the last. The list holds one more TERM than that, because the first term is there before any step is taken: $n = 24 + 1 = 25$. Dropping the $+1$ undercounts every finite AP by exactly one, which is why the two candidate answers here differ by a single term.

The common difference is defined as $a_{k+1} - a_k$: the term you arrive at minus the term you left. Here $5 - 8 = -3$, and every later pair gives the same. Subtracting the other way round reports $+3$, which says the list climbs when it plainly falls. The sign is not decoration -- every use of $a$ and $d$ after this inherits it, so an $n$th term computed from $+3$ runs off in the wrong direction entirely.

The squares of $3$, $4$, $5$, $6$ follow a rule anyone can continue, and are not an AP: consecutive squares differ by $7$ then $9$, and that gap grows for ever. The list that adds $5$ each time is the AP. The remaining list opens with a difference of $4$ and then changes, so trusting the first pair picks it. Neither being predictable nor starting well makes a list arithmetic; only a constant difference does.

First the general term: $a_n = S_n - S_{n-1} = 6n - 3 + 5$, so $a_{ 10 } = 62$. Two answers on offer are the right size but the wrong quantity. Subtracting the other way round gives $-62$ -- the same magnitude with the sign turned over, which no check on units will catch. And $S_{ 10 } = 350$ is the total of all $10$ terms, not the $10$th of them; it is far larger, and its size is the only clue that it answers a different question.

$a_n$ is what the running total GAINS when the $n$th term is added, so $a_n = S_n - S_{n-1} = (3n^2 + 5n) - [3(n-1)^2 + 5(n-1)]$ = $6n + 2$. Lowering the degree of $S_n$ by inspection gives $3n + 5$, which is a different function at every single $n$ -- the two never agree, so the shortcut is not even occasionally right. Notice what the answer confirms: $a_n$ came out LINEAR in $n$, which is what makes this list an AP at all, and its coefficient $6$ is the common difference.

The list FALLS, so $d = -7$ and the minus sign must be substituted into the formula: $a_{ 21 } = 28 + (21 - 1)(-7) = -112$. Using $7$ instead of $-7$ gives $168$, a positive number instead of the correct negative one. Taking $21$ steps instead of $20$ adds one extra common difference.

The reading does not change when the coefficient is irrational: whatever multiplies $n$ is the common difference, so $d = \sqrt{ 5 }$. Nothing in the definition requires $d$ to be rational -- the chapter's own $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, \dots$ has $d = \sqrt{2}$. What $d$ must be is a single fixed NUMBER, so $\sqrt{ 5 }\,n$ is not a possible answer however the formula is written, and the constant $1$ is $a - d$ rather than $d$.

The list falls, so $d = -4$ and $S = \frac{ 22 }{2}[24 - 21(4)] = -660$ -- a negative total, because the terms go negative and stay there. At $a = 8$, $d = -5$ and $n = 22$ this is the chapter's own Example 11 and the answer is $-979$. Both wrong routes survive the sign change unharmed: halving the leading term gives $-792$, and taking $22$ steps instead of $21$ gives $-704$.

Here $d = -2$, and the quadratic $2n^2 - 34n + 104 = 0$ has roots $4$ and $13$, both whole and positive. The AP passes through zero and goes negative, so the terms from the $5$th to the $13$th cancel out exactly. Test each root against the situation instead of discarding the larger one. At $a = 24$, $d = -3$ and $S = 78$ this is the board's own question, whose answer is $4$ and $13$.

The first term never enters: subtracting two terms cancels it and leaves $(24 - 19)d = 5d$. So $d = 30 \div 5 = 6$. Handing back $30$ is the commonest slip in the whole chapter -- it reports $5$ steps' worth of change as one step's worth. Dividing by $6$ instead of $5$ is the other route, and it counts the two named terms as well as the steps between them.

Between the two dates $7$ increments were paid, and the salary rose by Rs $3500$, so each increment is $3500 \div 7 =$ Rs $500$. The whole rise is not the increment -- it is $7$ of them. Working back, the starting salary is $15500 - 5 \times 500 =$ Rs $13000$, since "after 5 years of service" means 5 increments have already been paid.

Consecutive terms of an AP have equal successive differences. The difference is $2 - -2 = 4$, so $y = 2 + 4 = 6$. Taking the difference the wrong way, $-2 - 2$, and adding it to $2$ gives $-2$.

Write the terms as $m - d$, $m$, $m + d$. Their sum is $3m = 12$, so $m = 4$. With common difference $3$, the numbers are 1, 4, 7. Treating the sum as three times the first term instead gives 4, 7, 10.