Let a_1, a_2, a_3, \ldots be in an arithmetic progression of positive terms. Let A_k=a_1^2-a_2^2+a_3^2-a_4^2+\ldots+a_2 k-1^2-a_2 k^2 . If A_3=-153, ~A_5=-435 and a_1^2+a_2^2+a_3^2=66 , then a_17-A_7 is equal to \underline\hspace2cm .

Sequences and Series PYQ — JEE Main 2024 · 5 Apr · Shift 1 · Q57 with Solution

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Correct answer: 910

Let the arithmetic progression be $a_n=a+(n-1)d,$ where $a=a_1$ and $d$ is the common difference.

We are given $A_k=a_1^2-a_2^2+a_3^2-a_4^2+ rac{}{}\cdots+a_{2k-1}^2-a_{2k}^2.$

We must find $a_{17}-A_7$ .

1. Simplify $A_k$

Group terms in pairs: $A_k=(a_1^2-a_2^2)+(a_3^2-a_4^2)+\cdots+(a_{2k-1}^2-a_{2k}^2).$

Now,

\qquad a_{2r}=a+(2r-1)d.$$ So, $$a_{2r-1}^2-a_{2r}^2=(a_{2r-1}-a_{2r})(a_{2r-1}+a_{2r}).$$ Since $$a_{2r-1}-a_{2r}=-d,$$ and $$a_{2r-1}+a_{2r}=[a+(2r-2)d]+[a+(2r-1)d]=2a+(4r-3)d,$$ we get $$a_{2r-1}^2-a_{2r}^2=-d\,[2a+(4r-3)d].$$ Hence, $$A_k=-d\sum_{r=1}^k [2a+(4r-3)d].$$ Now, $$\sum_{r=1}^k [2a+(4r-3)d]=2ak+d\sum_{r=1}^k(4r-3).$$ And $$\sum_{r=1}^k(4r-3)=4\cdot\frac{k(k+1)}2-3k=2k(k+1)-3k=2k^2-k.$$ Therefore, $$A_k=-d\left(2ak+(2k^2-k)d\right).$$ So, $$\boxed{A_k=-2akd-(2k^2-k)d^2}. $$ --- ## 2. Use $A_3=-153$ and $A_5=-435$ For $k=3$: $$A_3=-6ad-15d^2=-153$$ $$\Rightarrow 6ad+15d^2=153. \tag{1}$$ For $k=5$: $$A_5=-10ad-45d^2=-435$$ $$\Rightarrow 10ad+45d^2=435. \tag{2}$$ Solve (1) and (2). Multiply (1) by $5$: $$30ad+75d^2=765. \tag{3}$$ Multiply (2) by $3$: $$30ad+135d^2=1305. \tag{4}$$ Subtract (3) from (4): $$60d^2=540$$ $$d^2=9.$$ Since all terms are positive and later we will see $a>0$, take possibilities $d=\pm 3$ for now. From (1): $$6ad+15(9)=153$$ $$6ad+135=153$$ $$6ad=18$$ $$ad=3.$$ Thus, $$d^2=9,\qquad ad=3.$$ Hence: - if $d=3$, then $a=1$; - if $d=-3$, then $a=-1$. But terms are positive, and $a_1=a$ must be positive, so $a=-1$ is impossible. Therefore, $$\boxed{a=1,\ d=3}. $$ So the AP is $$1,4,7,10,\dots$$ --- ## 3. Verify with $a_1^2+a_2^2+a_3^2=66$ We check: $$a_1=1,\ a_2=4,\ a_3=7.$$ Then $$a_1^2+a_2^2+a_3^2=1^2+4^2+7^2=1+16+49=66,$$ which matches the given condition. --- ## 4. Find $a_{17}$ $$a_{17}=a+16d=1+16\cdot 3=49.$$ So, $$\boxed{a_{17}=49}. $$ --- ## 5. Find $A_7$ Using $$A_k=-2akd-(2k^2-k)d^2,$$ with $a=1$, $d=3$, $k=7$: $$A_7=-2(1)(7)(3)-(2\cdot 7^2-7)(9).$$ Now, $$-2(1)(7)(3)=-42,$$ and $$2\cdot 7^2-7=98-7=91.$$ So, $$A_7=-42-91\cdot 9=-42-819=-861.$$ Thus, $$\boxed{A_7=-861}. $$ --- ## 6. Compute $a_{17}-A_7$ $$a_{17}-A_7=49-(-861)=49+861=910.$$ Therefore, $$\boxed{910}$$ --- ## 7. Comparison with stored answer Stored correct answer: $910$ Our derived answer is also $910$, so it agrees.

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1. Simplify

A_k

Group terms in pairs:

A_k=(a_1^2-a_2^2)+(a_3^2-a_4^2)+\cdots+(a_{2k-1}^2-a_{2k}^2).

Now,

\qquad a_{2r}=a+(2r-1)d.$$ So, $$a_{2r-1}^2-a_{2r}^2=(a_{2r-1}-a_{2r})(a_{2r-1}+a_{2r}).$$ Since $$a_{2r-1}-a_{2r}=-d,$$ and $$a_{2r-1}+a_{2r}=[a+(2r-2)d]+[a+(2r-1)d]=2a+(4r-3)d,$$ we get $$a_{2r-1}^2-a_{2r}^2=-d\,[2a+(4r-3)d].$$ Hence, $$A_k=-d\sum_{r=1}^k [2a+(4r-3)d].$$ Now, $$\sum_{r=1}^k [2a+(4r-3)d]=2ak+d\sum_{r=1}^k(4r-3).$$ And $$\sum_{r=1}^k(4r-3)=4\cdot\frac{k(k+1)}2-3k=2k(k+1)-3k=2k^2-k.$$ Therefore, $$A_k=-d\left(2ak+(2k^2-k)d\right).$$ So, $$\boxed{A_k=-2akd-(2k^2-k)d^2}. $$ --- ## 2. Use $A_3=-153$ and $A_5=-435$ For $k=3$: $$A_3=-6ad-15d^2=-153$$ $$\Rightarrow 6ad+15d^2=153. \tag{1}$$ For $k=5$: $$A_5=-10ad-45d^2=-435$$ $$\Rightarrow 10ad+45d^2=435. \tag{2}$$ Solve (1) and (2). Multiply (1) by $5$: $$30ad+75d^2=765. \tag{3}$$ Multiply (2) by $3$: $$30ad+135d^2=1305. \tag{4}$$ Subtract (3) from (4): $$60d^2=540$$ $$d^2=9.$$ Since all terms are positive and later we will see $a>0$, take possibilities $d=\pm 3$ for now. From (1): $$6ad+15(9)=153$$ $$6ad+135=153$$ $$6ad=18$$ $$ad=3.$$ Thus, $$d^2=9,\qquad ad=3.$$ Hence: - if $d=3$, then $a=1$; - if $d=-3$, then $a=-1$. But terms are positive, and $a_1=a$ must be positive, so $a=-1$ is impossible. Therefore, $$\boxed{a=1,\ d=3}. $$ So the AP is $$1,4,7,10,\dots$$ --- ## 3. Verify with $a_1^2+a_2^2+a_3^2=66$ We check: $$a_1=1,\ a_2=4,\ a_3=7.$$ Then $$a_1^2+a_2^2+a_3^2=1^2+4^2+7^2=1+16+49=66,$$ which matches the given condition. --- ## 4. Find $a_{17}$ $$a_{17}=a+16d=1+16\cdot 3=49.$$ So, $$\boxed{a_{17}=49}. $$ --- ## 5. Find $A_7$ Using $$A_k=-2akd-(2k^2-k)d^2,$$ with $a=1$, $d=3$, $k=7$: $$A_7=-2(1)(7)(3)-(2\cdot 7^2-7)(9).$$ Now, $$-2(1)(7)(3)=-42,$$ and $$2\cdot 7^2-7=98-7=91.$$ So, $$A_7=-42-91\cdot 9=-42-819=-861.$$ Thus, $$\boxed{A_7=-861}. $$ --- ## 6. Compute $a_{17}-A_7$ $$a_{17}-A_7=49-(-861)=49+861=910.$$ Therefore, $$\boxed{910}$$ --- ## 7. Comparison with stored answer Stored correct answer: $910$ Our derived answer is also $910$, so it agrees.

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