The 2nd 3rd and 14th term of an AP are 3concecutive term ofaGP the 10ty term of AP is 18 find first term and common different of the progration

The 2nd term of the AP is $a_2 = a + (2-1)d = a + d$.
The 3rd term of the AP is $a_3 = a + (3-1)d = a + 2d$.
The 14th term of the AP is $a_{14} = a + (14-1)d = a + 13d$.

If $a_2, a_3, a_{14}$ are consecutive terms of a GP, then the ratio of consecutive terms is constant.

This implies $(a_3)^2 = a_2 \cdot a_{14}$.
Substitute the expressions from Step 1:

Expand both sides:

Subtract $a^2$ from both sides:

Rearrange the terms to one side:

Factor out $d$:

This equation gives two possibilities for $d$:
Possibility 1: $d = 0$
Possibility 2: $10a + 9d = 0$

The 10th term of the AP is 18.

So, we have the equation:

• Case 1: $d = 0$ Substitute $d=0$ into the equation $a + 9d = 18$:

  $$a + 9(0) = 18$$ $$a = 18$$

  In this case, the first term is 18 and the common difference is 0.

• Case 2: $10a + 9d = 0$ From this equation, express $a$ in terms of $d$:

  $$10a = -9d$$ $$a = -\frac{9}{10}d$$

  Now substitute this expression for $a$ into the equation $a + 9d = 18$:

  $$-\frac{9}{10}d + 9d = 18$$

  To eliminate the fraction, multiply the entire equation by 10:

  $$-9d + 90d = 180$$ $$81d = 180$$ $$d = \frac{180}{81}$$

  Simplify the fraction by dividing both numerator and denominator by 9:

  $$d = \frac{20}{9}$$

  Now find $a$ using $a = -\frac{9}{10}d$:

  $$a = -\frac{9}{10} \cdot \frac{20}{9}$$ $$a = -\frac{180}{90}$$ $$a = -2$$

  In this case, the first term is -2 and the common difference is $\frac{20}{9}$.

There are two possible pairs for the first term and common difference of the progression.

The first term and common difference are:

• $a = 18$, $d = 0$
• $a = -2$, $d = \frac{20}{9}$