Let aₙ be the nᵗʰ term of an A.P. If Sₙ = a₁ + a₂ + a₃ + ⋯ + aₙ = 700, a₆ = 7 and S₇ = 7, then aₙ is equal to:

❓ Question

Let $a_n$ be the $n^{\text{th}}$ term of an A.P. If $S_n=a_1+a_2+\cdots+a_n=700,\; a_6=7$ and $S_7=7$, then find $a_n$.

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🖼️ Question Image

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✍️ Short Solution

Let the A.P. have first term $a_1$ and common difference $d$. Then:

• $a_6=a_1+5d=7.$

• Sum of first 7 terms:

  $$S_7 = \frac{7}{2}\big(2a_1 + 6d\big) = 7.$$
  

  Divide both sides by 7:

  $$\frac{1}{2}\big(2a_1 + 6d\big) = 1 \quad\Rightarrow\quad a_1 + 3d = 1.$$
  

Now subtract the two linear equations:

$$(a_1+5d) - (a_1+3d) = 7 - 1 \;\Rightarrow\; 2d = 6 \;\Rightarrow\; d = 3.$$

Then $a_1 + 3d = 1 \Rightarrow a_1 + 9 = 1 \Rightarrow a_1 = -8.$

So the A.P. is $-8,-5,-2,1,4,7,10,\dots$

Next, use the condition $S_n = 700$. Sum formula:

$$S_n=\frac{n}{2}\big(2a_1+(n-1)d\big)=\frac{n}{2}\big(2(-8)+(n-1)3\big) =\frac{n}{2}(3n-19).$$

Hence

$$\frac{n(3n-19)}{2}=700 \quad\Rightarrow\quad n(3n-19)=1400.$$

This gives the quadratic:

$$3n^2-19n-1400=0.$$

Compute discriminant: $\Delta = (-19)^2 + 4\cdot3\cdot1400 = 361 + 16800 = 17161 =131^2.$
So positive root:

$$n=\frac{19+131}{6}=\frac{150}{6}=25.$$

Therefore $n=25$. Finally,

$$a_n = a_1 + (n-1)d = -8 + 24\cdot 3 = -8 + 72 = 64.$$

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✅ Conclusion & Video Solution