Let a_1, a_2, a_3, be an A.P.

Question

Let $a_1, a_2, a_3, \ldots$ be an A.P. and $g_1 = a_1, g_2, g_3, \ldots$ be an increasing G.P. If $a_1 = a_2 + g_2 = 1$ and $a_3 + g_3 = 4$, then $a_{10} + g_5$ is equal to:

Accepted Answer

Let the common difference of the given A.P. be $d$ and the common ratio of the given G.P. be $r$.

From the general formulas for the $n$-th term of an A.P. and a G.P., we know:

${a}_{n}={a}_{1}+(n-1)d$

${g}_{n}={g}_{1}{r}^{n-1}$

We are given that ${g}_{1}={a}_{1}$ and ${a}_{1}=1$.

Thus, the first term of both the A.P. and the G.P. is $1$. So, ${g}_{1}=1$.

Next, we are given the first relationship:

${a}_{2}+{g}_{2}=1$

Using our formulas, we can write the second terms as ${a}_{2}=1+d$ and ${g}_{2}=1\cdot r=r$. Let us substitute these values into the given equation:

$(1+d)+r=1$

Subtracting $1$ from both sides, we establish a relationship between $d$ and $r$:

$d+r=0\, \implies \, d=-r$

We are also given a second relationship:

${a}_{3}+{g}_{3}=4$

Again, applying the $n$-th term formulas, we have the third terms as ${a}_{3}=1+2d$ and ${g}_{3}=1\cdot {r}^{2}={r}^{2}$. Substituting these in the equation gives:

$(1+2d)+{r}^{2}=4$

Now, let us substitute $d=-r$ into this equation to form a quadratic equation in terms of $r$:

$1+2(-r)+{r}^{2}=4$

${r}^{2}-2r+1-4=0$

${r}^{2}-2r-3=0$

Let us factorize this quadratic equation by splitting the middle term:

${r}^{2}-3r+r-3=0$

$r(r-3)+1(r-3)=0$

$(r-3)(r+1)=0$

This gives us two possible values for $r$: $r=3$ or $r=-1$.

The problem clearly states that the G.P. is an increasing sequence. Since the first term ${g}_{1}=1$ is positive, an increasing sequence requires the common ratio to be strictly greater than $1$. Therefore, we reject $r=-1$ and take the valid value:

$r=3$

Since $d=-r$, we find the common difference of the A.P.:

$d=-3$

Now, we are asked to find the value of ${a}_{10}+{g}_{5}$. Let us find ${a}_{10}$ and ${g}_{5}$ individually.

For the 10th term of the A.P.:

${a}_{10}={a}_{1}+(10-1)d$

${a}_{10}={a}_{1}+9d$

${a}_{10}=1+9(-3)$

${a}_{10}=1-27=-26$

For the 5th term of the G.P.:

${g}_{5}={g}_{1}{r}^{5-1}$

${g}_{5}={g}_{1}{r}^{4}$

${g}_{5}=1\cdot (3{)}^{4}$

${g}_{5}=81$

Finally, adding these two values together:

${a}_{10}+{g}_{5}=-26+81=55$

Therefore, the value of ${a}_{10}+{g}_{5}$ is $55$.

Correct Answer:

Option D: $55$

Solutions · 1